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भाग 1 - सामान्य परिचय - अध्याय 6 - बल गति को कैसे प्रभावित करते हैं - पेज 94-99
सामान्य परिचय पेज 94: अध्याय 4 में स्थिति वेग त्वरण सीखा लेकिन कारण क्या गति बदलने का? इस अध्याय में बल क्या गति कैसे प्रभावित करता न्यूटन तीन नियम सीखेंगे। Think It Over कैनो आगे क्यों चलती जब कैनोइस्ट पानी पीछे धकेलता? तेज धकेलने पर तेज क्यों? खाली और सवारी वाली में कौन तेज? - खाली द्रव्यमान कम a=F/m अधिक तेज Fig6.28
बल क्या पेज 95 Fig6.1: गेंद लात विराम से गति Fig6.1a, बल्ले से दिशा बदलना Fig6.1b, नींबू निचोड़ना आकार बदलना Fig6.1c - बल परिमाण दिशा दोनों - SI न्यूटन N - स्प्रिंग तुला Fig6.2 - 10⁻³ N millinewton 10⁻²⁴ N yoctonewton 2026 तक
संतुलित असंतुलित पेज 96 Example6.1: बक्सा धकेलना घर्षण विपरीत Fig6.3a, पानी गेंद गुरुत्व नीचे उत्प्लावन ऊपर Fig6.3b - रस्साकशी बराबर बल रस्सी नहीं हिलती Fig6.4a संतुलित, एक टीम अधिक बल रस्सी बड़े बल दिशा Fig6.4b असंतुलित नेट अंतर दिशा बड़े बल की, same दिशा दो लोग गाड़ी Fig6.5 नेट योग - Example6.1 10N 6N (a)16N दाएँ (b)4N दाएँ (c)4N बाएँ
बल क्या पेज 95 Fig6.1: गेंद लात विराम से गति Fig6.1a, बल्ले से दिशा बदलना Fig6.1b, नींबू निचोड़ना आकार बदलना Fig6.1c - बल परिमाण दिशा दोनों - SI न्यूटन N - स्प्रिंग तुला Fig6.2 - 10⁻³ N millinewton 10⁻²⁴ N yoctonewton 2026 तक
संतुलित असंतुलित पेज 96 Example6.1: बक्सा धकेलना घर्षण विपरीत Fig6.3a, पानी गेंद गुरुत्व नीचे उत्प्लावन ऊपर Fig6.3b - रस्साकशी बराबर बल रस्सी नहीं हिलती Fig6.4a संतुलित, एक टीम अधिक बल रस्सी बड़े बल दिशा Fig6.4b असंतुलित नेट अंतर दिशा बड़े बल की, same दिशा दो लोग गाड़ी Fig6.5 नेट योग - Example6.1 10N 6N (a)16N दाएँ (b)4N दाएँ (c)4N बाएँ
घर्षण पेज 97-99 Activity6.1: बक्सा फर्श बीच घर्षण विपरीत Fig6.10 - आपका बल घर्षण से बड़ा तभी हिलेगा नेट गति दिशा - गुरुत्व नीचे अभिलंब ऊपर संतुलित Fig6.11 हवा घर्षण छोटा नगण्य - Activity6.1 चार सिक्के रबर बैंड A B C लकड़ी सीमेंट पॉलिश संगमरमर दूरी घर्षण कम वेग धीरे घटा दूरी अधिक - विचार प्रयोग घर्षण शून्य कभी नहीं रुकेगा हमेशा चलता रहेगा - गैलीलियो विचार - Pause Q1 Q2 Fig6.8 Fig6.9
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भाग 2 - परिभाषा सिद्धांत सूत्र - पेज 100-106 पूरे पद Eq6.1 Eq6.2 Eq6.3
फ्रेम 1/4 - पहला नियम परिभाषा सिद्धांत पेज100 6.4 पूरे पद
Ancient times: It was thought that a force is needed to maintain constant velocity for centuries this incorrect idea continued. In 17th century Galileo Galilei thought experiment - On a horizontal plane if all obstacles removed object would continue moving indefinitely - Newton called this property inertia tendency of object to resist change in state of rest or uniform motion.
Newton’s First Law परिभाषा पेज100: "An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object."
If net force zero it will not start moving and its velocity will not change acceleration zero. Rest means velocity zero constant velocity means magnitude and direction does not change moves in straight line same direction.
Net zero graphs पेज101 Fig6.15 Fig6.16 Example6.2 6.3 पूरे पद:
Example6.2: A box is being pushed forward with a force and friction equal opposite so balanced net zero will it continue moving with constant velocity? Yes if already moving.
Example6.3: If net zero two possibilities - either at rest or moving with constant velocity.
Fig6.15 At rest position does not change with time Fig6.15a velocity zero Fig6.15b.
Fig6.16 Constant velocity same velocity Fig6.16b horizontal position-time straight line Fig6.16a.
Note: If friction zero to move object at rest net needed, once moving with constant velocity no force needed to maintain, to change velocity magnitude direction or to stop force needed.
Q3 If object moving with constant velocity is net acting? No net zero.
Q4 (i) An object at rest will remain at rest (ii) Moving object will continue moving with constant velocity (iii) With constant acceleration - (i)(ii) possible (iii) not possible when net zero.
Q5 In real world no force difficult extra force to make net zero box friction equal net zero constant velocity.
Ancient times: It was thought that a force is needed to maintain constant velocity for centuries this incorrect idea continued. In 17th century Galileo Galilei thought experiment - On a horizontal plane if all obstacles removed object would continue moving indefinitely - Newton called this property inertia tendency of object to resist change in state of rest or uniform motion.
Newton’s First Law परिभाषा पेज100: "An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object."
If net force zero it will not start moving and its velocity will not change acceleration zero. Rest means velocity zero constant velocity means magnitude and direction does not change moves in straight line same direction.
Net zero graphs पेज101 Fig6.15 Fig6.16 Example6.2 6.3 पूरे पद:
Example6.2: A box is being pushed forward with a force and friction equal opposite so balanced net zero will it continue moving with constant velocity? Yes if already moving.
Example6.3: If net zero two possibilities - either at rest or moving with constant velocity.
Fig6.15 At rest position does not change with time Fig6.15a velocity zero Fig6.15b.
Fig6.16 Constant velocity same velocity Fig6.16b horizontal position-time straight line Fig6.16a.
Note: If friction zero to move object at rest net needed, once moving with constant velocity no force needed to maintain, to change velocity magnitude direction or to stop force needed.
Q3 If object moving with constant velocity is net acting? No net zero.
Q4 (i) An object at rest will remain at rest (ii) Moving object will continue moving with constant velocity (iii) With constant acceleration - (i)(ii) possible (iii) not possible when net zero.
Q5 In real world no force difficult extra force to make net zero box friction equal net zero constant velocity.
फ्रेम 2/4 - दूसरा नियम परिभाषा सिद्धांत सूत्र पेज102-104 Eq6.1 Eq6.2 Eq6.3 पूरे पद
6.5 Newton’s Second Law of Motion पेज102: You know force can set object in motion bring to rest or change velocity Change in velocity means object accelerating Thus force produces acceleration But what is relationship between net force and acceleration? From everyday experiences if ball pushed gently moves slowly starting from rest i.e. acceleration small On other hand strong push ball starts to move fast i.e. larger acceleration So hypothesis for same object larger force results in larger acceleration smaller force smaller acceleration. Now how test? Activity where apply forces different magnitudes on same object same surface to find resultant acceleration How apply forces different magnitudes? You have learnt gravitational force weight different for different objects Hence you can use weights different magnitudes to apply forces different magnitudes.
Activity6.3 Let us experiment पेज102 Fig6.17: Take four ball bearing wheels two pencils empty cardboard box to make cart paper cup piece of pipe to use as pulley length thread some coins weighing scale. Insert two pencils through sides near bottom to function as axles attach wheel to each free ends Fig6.17a. Attach thread to front end. Draw line at one end table marking starting point. Put thread over small pipe at other end Fig6.17b To thread attach cup in which you can put objects As you let system go cup will move down due to gravitational force by Earth and thread will pull cart with constant force. Measure mass of cup along with objects with weighing scale. Start recording video cart in slow motion Release cart from start line record until reaches pipe at other end. Fig6.17 (a) Cart (b) cart and pulley system.
Result same object larger force larger acceleration. Same force lighter easier heavier difficult smaller mass larger acceleration Activity6.4 mass doubled same force acceleration reduced.
Second Law परिभाषा: "When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object."
सूत्र Eq6.1 Eq6.2 Eq6.3 पेज104: Mathematically a = F/m (6.1) or F=ma (6.2) where a acceleration F force m mass direction of acceleration same as net force. SI units mass kg acceleration m s⁻² If m=1kg a=1 m s⁻² then using Eq6.2 F=1kg×1 m s⁻² =1 kg m s⁻² =1 N One newton defined as force that produces acceleration 1 m s⁻² on object mass 1kg. Under gravitational influence acceleration due to gravitational force by Earth denoted g unit same as acceleration m s⁻² Using Eq6.2 gravitational force F=mg (6.3) Value g=9.8 m s⁻² nearly constant near surface For quick estimations g=10 m s⁻². Momentum product mass velocity direction same as velocity Rate change momentum proportional to net force direction net force applicable even mass not constant.
Note g does not depend on mass, 100g mass in palm upward force around 1N.
6.5 Newton’s Second Law of Motion पेज102: You know force can set object in motion bring to rest or change velocity Change in velocity means object accelerating Thus force produces acceleration But what is relationship between net force and acceleration? From everyday experiences if ball pushed gently moves slowly starting from rest i.e. acceleration small On other hand strong push ball starts to move fast i.e. larger acceleration So hypothesis for same object larger force results in larger acceleration smaller force smaller acceleration. Now how test? Activity where apply forces different magnitudes on same object same surface to find resultant acceleration How apply forces different magnitudes? You have learnt gravitational force weight different for different objects Hence you can use weights different magnitudes to apply forces different magnitudes.
Activity6.3 Let us experiment पेज102 Fig6.17: Take four ball bearing wheels two pencils empty cardboard box to make cart paper cup piece of pipe to use as pulley length thread some coins weighing scale. Insert two pencils through sides near bottom to function as axles attach wheel to each free ends Fig6.17a. Attach thread to front end. Draw line at one end table marking starting point. Put thread over small pipe at other end Fig6.17b To thread attach cup in which you can put objects As you let system go cup will move down due to gravitational force by Earth and thread will pull cart with constant force. Measure mass of cup along with objects with weighing scale. Start recording video cart in slow motion Release cart from start line record until reaches pipe at other end. Fig6.17 (a) Cart (b) cart and pulley system.
Result same object larger force larger acceleration. Same force lighter easier heavier difficult smaller mass larger acceleration Activity6.4 mass doubled same force acceleration reduced.
Second Law परिभाषा: "When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object."
सूत्र Eq6.1 Eq6.2 Eq6.3 पेज104: Mathematically a = F/m (6.1) or F=ma (6.2) where a acceleration F force m mass direction of acceleration same as net force. SI units mass kg acceleration m s⁻² If m=1kg a=1 m s⁻² then using Eq6.2 F=1kg×1 m s⁻² =1 kg m s⁻² =1 N One newton defined as force that produces acceleration 1 m s⁻² on object mass 1kg. Under gravitational influence acceleration due to gravitational force by Earth denoted g unit same as acceleration m s⁻² Using Eq6.2 gravitational force F=mg (6.3) Value g=9.8 m s⁻² nearly constant near surface For quick estimations g=10 m s⁻². Momentum product mass velocity direction same as velocity Rate change momentum proportional to net force direction net force applicable even mass not constant.
Note g does not depend on mass, 100g mass in palm upward force around 1N.
फ्रेम 3/4 - अनुप्रयोग पेज104-105 Fig6.18 Fig6.19 Fig6.20 पूरे पद
Bridging Science and Society: In game of cricket you might have noticed while catching fast-moving ball fielder gradually pulls hands backwards with moving ball just after catching Fig6.18 In doing so time duration increased during which high velocity reduces to zero This reduces magnitude of acceleration of ball as it slows down thus requiring smaller force to be applied by fielder to stop ball Applying smaller force to moving ball also minimises injury to fielder Fig6.18 Catching cricket ball.
For similar reason airbags provided in vehicles Fig6.19 In event of collision vehicle coming to abrupt halt airbag inflates quickly into soft compressible cushion Instead of directly hitting hard steering wheel or dashboard passenger’s head and chest push into bag and time over which hitting occurs increases Smaller acceleration means reduced force exerted on person thereby lowering risk serious injuries particularly when combined with seat belt usage Fig6.19 Inflated airbag.
Have you ever attempted cracking fresh coconut in one go? It is brought down at very high velocity to hit hard surface Fig6.20 When coconut hits ground it stops in very short time To change its velocity so quickly ground must exert very large force on it This large force breaks shell Fig6.20 Cracking coconut.
Threads of Curiosity: As per Eq6.1 in Activity6.3 doubling force should double acceleration but reality increase little less than factor two Similarly Activity6.4 doubled mass acceleration should halved but slightly different Apart from measurement errors friction between cart wheels and surface can lead differences.
Bridging Science and Society: In game of cricket you might have noticed while catching fast-moving ball fielder gradually pulls hands backwards with moving ball just after catching Fig6.18 In doing so time duration increased during which high velocity reduces to zero This reduces magnitude of acceleration of ball as it slows down thus requiring smaller force to be applied by fielder to stop ball Applying smaller force to moving ball also minimises injury to fielder Fig6.18 Catching cricket ball.
For similar reason airbags provided in vehicles Fig6.19 In event of collision vehicle coming to abrupt halt airbag inflates quickly into soft compressible cushion Instead of directly hitting hard steering wheel or dashboard passenger’s head and chest push into bag and time over which hitting occurs increases Smaller acceleration means reduced force exerted on person thereby lowering risk serious injuries particularly when combined with seat belt usage Fig6.19 Inflated airbag.
Have you ever attempted cracking fresh coconut in one go? It is brought down at very high velocity to hit hard surface Fig6.20 When coconut hits ground it stops in very short time To change its velocity so quickly ground must exert very large force on it This large force breaks shell Fig6.20 Cracking coconut.
Threads of Curiosity: As per Eq6.1 in Activity6.3 doubling force should double acceleration but reality increase little less than factor two Similarly Activity6.4 doubled mass acceleration should halved but slightly different Apart from measurement errors friction between cart wheels and surface can lead differences.
फ्रेम 4/4 - Example6.4 6.5 6.6 और Pause Q6 Q7 Q8 पूरे पद पेज105-106 Fig6.21
Example6.4 पेज105 Fig6.8: A weight lifter is holding a barbell with mass of 10 kg fixed on each side of bar Fig6.8 Mass of bar itself is 10 kg How much force is she applying to keep barbell steady? Answer Total mass barbell 30 kg Gravitational force due to Earth acting downward using Eq6.3 F=mg=30kg×9.8 m s⁻²=294 N To keep steady weightlifter has to apply equal force opposite direction So she applying 294 N upward.
Example6.5 पेज105: A student trying to push stationary block of 25 kg on horizontal floor Maximum force friction opposing motion 50 N Determine displacement in 2 seconds if Rahul pushes with constant force (i)50 N and (ii)55 N forward. Answer (i) Force equal opposing friction balanced net zero block remains stationary (ii) Net 55-50=5 N Mass 25 kg Using Newton second law acceleration =F/m=5/25=0.2 m s⁻² Using kinematic s=ut+½at² =0×2+½×0.2×4=0.4 m forward.
Example6.6 पेज106 Fig6.21: A sports car mass 1500 kg moving towards east and velocity-time graph Fig6.21 Calculate force during (i)0-5s (ii)5-10s (iii)10-15s Answer (i) During 0-5s velocity-time straight inclined time axis indicates constant acceleration u=0 v=10 t=5 Using v=u+at 10=0+a×5 a=2 m s⁻² Using F=ma F=1500×2=3000 N towards east (ii) During 5-10s straight parallel time axis constant velocity hence no force (iii) During 10-15s straight inclined u=10 v=0 t=5 0=10+a×5 a=-2 F=1500×-2=-3000 N negative shows force opposite direction motion towards west Fig6.21 Velocity-time graph sports car.
Pause and Ponder: Q6 A toy car mass 100 g moving with constant velocity 0.5 m s⁻¹ What is net force? Answer zero constant velocity net zero first law.
Q7 Two children different masses sitting identical swings To impart identical initial acceleration for which child would you require larger force? Explain why. Answer larger mass larger force F=ma.
Q8 How are glass items packed for transportation using bubble wrap or hay protected from damage? Answer time increased acceleration reduced force reduced same as airbag coconut.
Example6.4 पेज105 Fig6.8: A weight lifter is holding a barbell with mass of 10 kg fixed on each side of bar Fig6.8 Mass of bar itself is 10 kg How much force is she applying to keep barbell steady? Answer Total mass barbell 30 kg Gravitational force due to Earth acting downward using Eq6.3 F=mg=30kg×9.8 m s⁻²=294 N To keep steady weightlifter has to apply equal force opposite direction So she applying 294 N upward.
Example6.5 पेज105: A student trying to push stationary block of 25 kg on horizontal floor Maximum force friction opposing motion 50 N Determine displacement in 2 seconds if Rahul pushes with constant force (i)50 N and (ii)55 N forward. Answer (i) Force equal opposing friction balanced net zero block remains stationary (ii) Net 55-50=5 N Mass 25 kg Using Newton second law acceleration =F/m=5/25=0.2 m s⁻² Using kinematic s=ut+½at² =0×2+½×0.2×4=0.4 m forward.
Example6.6 पेज106 Fig6.21: A sports car mass 1500 kg moving towards east and velocity-time graph Fig6.21 Calculate force during (i)0-5s (ii)5-10s (iii)10-15s Answer (i) During 0-5s velocity-time straight inclined time axis indicates constant acceleration u=0 v=10 t=5 Using v=u+at 10=0+a×5 a=2 m s⁻² Using F=ma F=1500×2=3000 N towards east (ii) During 5-10s straight parallel time axis constant velocity hence no force (iii) During 10-15s straight inclined u=10 v=0 t=5 0=10+a×5 a=-2 F=1500×-2=-3000 N negative shows force opposite direction motion towards west Fig6.21 Velocity-time graph sports car.
Pause and Ponder: Q6 A toy car mass 100 g moving with constant velocity 0.5 m s⁻¹ What is net force? Answer zero constant velocity net zero first law.
Q7 Two children different masses sitting identical swings To impart identical initial acceleration for which child would you require larger force? Explain why. Answer larger mass larger force F=ma.
Q8 How are glass items packed for transportation using bubble wrap or hay protected from damage? Answer time increased acceleration reduced force reduced same as airbag coconut.
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भाग 3 - अभ्यास - अध्याय 6 - प्रश्नावली के सभी प्रश्न पूरे पदों में - पेज 112-115
फ्रेम 1/4 - अभ्यास Q1 से Q4 पूरे पदों में [पेज 112-113 NCERT]
At a Glance पेज112:
• The force of friction acts on an object in the direction opposite to its direction of motion.
• Newton’s first law of motion: An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.
• Newton’s second law of motion: When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object.
• Newton’s third law of motion: Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object.
Q1. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
उत्तर: F के बराबर विपरीत दिशा में - क्योंकि नियत वेग नेट शून्य संतुलित Example6.2 जैसा पेज101
Q2. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
(i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.
(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
उत्तर: (i) remain the same (ii) increase (iii) decrease - पहला और दूसरा नियम पेज100-104
Q3. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity. Fig. 6.36 - 5 N → P ← 4 N (a) Q (b) Which of the following statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.
उत्तर: (i) सही - P पर नेट 5-4=1N, Q नियत वेग नेट0 पेज101 Example6.3
Q4. While practising for the snake boat race (Vallam kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
उत्तर: 95×200=19000 आगे, 5×200=1000 पीछे, नेट=19000-1000=18000 N आगे दिशा में - Fig6.5 जैसा योग अंतर पेज96
At a Glance पेज112:
• The force of friction acts on an object in the direction opposite to its direction of motion.
• Newton’s first law of motion: An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.
• Newton’s second law of motion: When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object.
• Newton’s third law of motion: Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object.
Q1. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
उत्तर: F के बराबर विपरीत दिशा में - क्योंकि नियत वेग नेट शून्य संतुलित Example6.2 जैसा पेज101
Q2. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
(i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.
(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
उत्तर: (i) remain the same (ii) increase (iii) decrease - पहला और दूसरा नियम पेज100-104
Q3. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity. Fig. 6.36 - 5 N → P ← 4 N (a) Q (b) Which of the following statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.
उत्तर: (i) सही - P पर नेट 5-4=1N, Q नियत वेग नेट0 पेज101 Example6.3
Q4. While practising for the snake boat race (Vallam kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
उत्तर: 95×200=19000 आगे, 5×200=1000 पीछे, नेट=19000-1000=18000 N आगे दिशा में - Fig6.5 जैसा योग अंतर पेज96
फ्रेम 2/4 - अभ्यास Q5 से Q8 पूरे पदों में [पेज 113 NCERT]
Q5. When a net force acts on an object, we observe that the object accelerates:
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.
उत्तर: (iv) सही - दूसरे नियम Eq6.2 F=ma दिशा नेट दिशा समानुपाती पेज102-104
Q6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
(i) Object A (सीधी ऊपर रेखा नियत वेग)
(ii) Object B (क्षैतिज रेखा विराम)
(iii) Object C (वक्र ऊपर बढ़ती ढाल त्वरण)
(iv) Object D (सीधी नीचे रेखा नियत वेग विपरीत)
Fig6.37 में A सीधी ऊपर, B क्षैतिज, C वक्र ऊपर, D सीधी नीचे
उत्तर: केवल Object C पर नेट बल - क्योंकि वक्र त्वरण दर्शाता है Fig6.15 Fig6.16 पेज101
Q7. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.
Fig6.38: A sailor jumping forward from small boat to shore
उत्तर: हाँ नाव पीछे जाएगी - तीसरा नियम - नाविक आगे बल लगाता नाव पर, नाव नाविक पर पीछे बराबर विपरीत बल - Fig6.23 पहिए कुर्सी जैसा पेज107-108
Q8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.
Fig6.39: A landing mat for a high jump event
उत्तर: समय बढ़ा त्वरण घटा बल कम चोट कम - क्रिकेट कैच हाथ पीछे Fig6.18 पेज104-105 और एयरबैग Fig6.19 पेज105 जैसा - दूसरा नियम F=ma समय बढ़ा a घटा
Q5. When a net force acts on an object, we observe that the object accelerates:
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.
उत्तर: (iv) सही - दूसरे नियम Eq6.2 F=ma दिशा नेट दिशा समानुपाती पेज102-104
Q6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
(i) Object A (सीधी ऊपर रेखा नियत वेग)
(ii) Object B (क्षैतिज रेखा विराम)
(iii) Object C (वक्र ऊपर बढ़ती ढाल त्वरण)
(iv) Object D (सीधी नीचे रेखा नियत वेग विपरीत)
Fig6.37 में A सीधी ऊपर, B क्षैतिज, C वक्र ऊपर, D सीधी नीचे
उत्तर: केवल Object C पर नेट बल - क्योंकि वक्र त्वरण दर्शाता है Fig6.15 Fig6.16 पेज101
Q7. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.
Fig6.38: A sailor jumping forward from small boat to shore
उत्तर: हाँ नाव पीछे जाएगी - तीसरा नियम - नाविक आगे बल लगाता नाव पर, नाव नाविक पर पीछे बराबर विपरीत बल - Fig6.23 पहिए कुर्सी जैसा पेज107-108
Q8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.
Fig6.39: A landing mat for a high jump event
उत्तर: समय बढ़ा त्वरण घटा बल कम चोट कम - क्रिकेट कैच हाथ पीछे Fig6.18 पेज104-105 और एयरबैग Fig6.19 पेज105 जैसा - दूसरा नियम F=ma समय बढ़ा a घटा
फ्रेम 3/4 - अभ्यास Q9 से Q12 पूरे पदों में [पेज 114 NCERT]
Q9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
उत्तर: (iv) सही - तीसरा नियम बल बराबर परिमाण विपरीत दिशा - द्रव्यमान अलग त्वरण अलग Example6.8 पेज110
Q10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
Fig6.40: Acceleration (m s⁻²) Y, Mass (kg) X - 1kg पर 10, 2kg पर 5, 5kg पर 2 - hyperbola a=F/m
उत्तर: F=ma = 1×10=10N, 2×5=10N, 5×2=10N - बल नियत 10N - force-mass ग्राफ क्षैतिज सीधी रेखा 10N पर - Eq6.2 पेज104
Q11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
Fig6.41: Velocity (m s⁻¹) Y 0-30, Time (s) X 0-8 - 0 से 4s तक 10 से 20, 4 से 8s तक 20 से 30 - सीधी रेखा ढाल
उत्तर: ढाल a = (30-10)/8 = 20/8=2.5 m/s², F=ma=10×2.5=25 N - Example6.6 जैसा Fig6.21 पेज106
Q12. A bullet of mass 50 g moving with a speed of 100 m s⁻¹ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).
उत्तर: m=0.05kg u=100 v=0 s=0.5m, v²=u²+2as, 0=10000+2a×0.5, a=-10000 m/s², F=ma=0.05×10000=500 N रोकने वाला बल - दूसरा नियम Eq6.2 पेज104
Q9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
उत्तर: (iv) सही - तीसरा नियम बल बराबर परिमाण विपरीत दिशा - द्रव्यमान अलग त्वरण अलग Example6.8 पेज110
Q10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
Fig6.40: Acceleration (m s⁻²) Y, Mass (kg) X - 1kg पर 10, 2kg पर 5, 5kg पर 2 - hyperbola a=F/m
उत्तर: F=ma = 1×10=10N, 2×5=10N, 5×2=10N - बल नियत 10N - force-mass ग्राफ क्षैतिज सीधी रेखा 10N पर - Eq6.2 पेज104
Q11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
Fig6.41: Velocity (m s⁻¹) Y 0-30, Time (s) X 0-8 - 0 से 4s तक 10 से 20, 4 से 8s तक 20 से 30 - सीधी रेखा ढाल
उत्तर: ढाल a = (30-10)/8 = 20/8=2.5 m/s², F=ma=10×2.5=25 N - Example6.6 जैसा Fig6.21 पेज106
Q12. A bullet of mass 50 g moving with a speed of 100 m s⁻¹ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).
उत्तर: m=0.05kg u=100 v=0 s=0.5m, v²=u²+2as, 0=10000+2a×0.5, a=-10000 m/s², F=ma=0.05×10000=500 N रोकने वाला बल - दूसरा नियम Eq6.2 पेज104
फ्रेम 4/4 - अभ्यास Q13 से Q16 पूरे पदों में [पेज 114 NCERT]
Q13. An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻¹. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
उत्तर: v=108km/h=30m/s, u=0, F=800N m=0.4kg, F×t = m(v-u) = 0.4×30=12, t=12/800=0.015 s - संवेग परिवर्तन दर = बल - दूसरा नियम पेज104 Note संवेग=mv
Q14. An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
उत्तर: घर्षण 7N + 3N =10N विपरीत नेट10N, a=F/m=10/2=5 m/s² विपरीत, u=10 v=0, v²=u²-2as, 0=100-2×5×s, s=10m - Example6.5 जैसा पेज105
Q15. A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.
उत्तर: F=m1a1, F=m2a2, कुल द्रव्यमान m1+m2, F=(m1+m2)a, a=F/(m1+m2), 1/a = m1/F + m2/F = 1/a1 +1/a2, a= a1a2/(a1+a2) - Eq6.4 a=F/(m1+m2) पेज111 प्रणाली Fig6.34
Q16. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
Fig6.42: A bar magnet and a magnetic compass
उत्तर: बल बराबर परिमाण विपरीत दिशा लेकिन द्रव्यमान अलग a=F/m अलग - सुई हल्की जड़त्व कम आसानी से चलती, बार चुंबक भारी जड़त्व अधिक - Example6.7 पृथ्वी फल Fig6.33 पेज110 और Example6.8 गोली बंदूक पेज110 जैसा - Note Even though forces equal, acceleration not equal because masses different पेज110
Q13. An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻¹. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
उत्तर: v=108km/h=30m/s, u=0, F=800N m=0.4kg, F×t = m(v-u) = 0.4×30=12, t=12/800=0.015 s - संवेग परिवर्तन दर = बल - दूसरा नियम पेज104 Note संवेग=mv
Q14. An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
उत्तर: घर्षण 7N + 3N =10N विपरीत नेट10N, a=F/m=10/2=5 m/s² विपरीत, u=10 v=0, v²=u²-2as, 0=100-2×5×s, s=10m - Example6.5 जैसा पेज105
Q15. A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.
उत्तर: F=m1a1, F=m2a2, कुल द्रव्यमान m1+m2, F=(m1+m2)a, a=F/(m1+m2), 1/a = m1/F + m2/F = 1/a1 +1/a2, a= a1a2/(a1+a2) - Eq6.4 a=F/(m1+m2) पेज111 प्रणाली Fig6.34
Q16. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
Fig6.42: A bar magnet and a magnetic compass
उत्तर: बल बराबर परिमाण विपरीत दिशा लेकिन द्रव्यमान अलग a=F/m अलग - सुई हल्की जड़त्व कम आसानी से चलती, बार चुंबक भारी जड़त्व अधिक - Example6.7 पृथ्वी फल Fig6.33 पेज110 और Example6.8 गोली बंदूक पेज110 जैसा - Note Even though forces equal, acceleration not equal because masses different पेज110
✅ भाग 3 - अभ्यास - सभी 16 प्रश्न पूरे पदों में - शॉर्ट नहीं - PDF पेज 112-115 से शुद्ध ✅
दांग से शिक्षा - बिगाड़ नहीं - शिक्षा का उन्मूलन नहीं
दांग से शिक्षा - बिगाड़ नहीं - शिक्षा का उन्मूलन नहीं
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भाग 4 - अग्निवीर - पेज 112-115 MCQ ट्रिक - पूरे पद - दांग से शिक्षा
फ्रेम 1/4 - अग्निवीर - Q4 से Q8 पूरे पद पेज113 Fig6.37 Fig6.38 Fig6.39
Q4. While practising for the snake boat race (Vallam kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
Ans: 95×200=19000 forward, 5×200=1000 backward, net 18000 N forward - Example6.1 Fig6.5
Q5. When a net force acts on an object, we observe that the object accelerates: (i) opposite to direction of force with acceleration proportional to force (ii) opposite to direction with proportional to mass (iii) in direction of force with acceleration inversely proportional to force (iv) in direction of force with acceleration proportional to force.
Ans: (iv) second law F=ma direction net force proportional to force p112
Q6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig.6.37. A net force acts on: (i) Object A (ii) Object B (iii) Object C (iv) Object D - Fig6.37 Position-Time - Object A straight up constant velocity, B horizontal rest, C curved increasing slope accelerating, D straight down constant velocity.
Ans: Only Object C - curved shows acceleration net needed Fig6.15 Fig6.16
Q7. A sailor jumps out from a small boat to the shore (Fig6.38). As sailor jumps forward, will boat move? If yes, in which direction and why. Fig6.38 A sailor jumping forward.
Ans: Yes boat moves backward third law sailor pushes boat backward boat pushes sailor forward Fig6.23
Q8. During high jump event landing mat or sand bed placed for athlete to fall upon Fig6.39 Explain reason. Fig6.39 Landing mat for high jump.
Ans: Time increased acceleration decreased force decreased injury minimized cricket catch Fig6.18 airbag Fig6.19
Q4. While practising for the snake boat race (Vallam kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
Ans: 95×200=19000 forward, 5×200=1000 backward, net 18000 N forward - Example6.1 Fig6.5
Q5. When a net force acts on an object, we observe that the object accelerates: (i) opposite to direction of force with acceleration proportional to force (ii) opposite to direction with proportional to mass (iii) in direction of force with acceleration inversely proportional to force (iv) in direction of force with acceleration proportional to force.
Ans: (iv) second law F=ma direction net force proportional to force p112
Q6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig.6.37. A net force acts on: (i) Object A (ii) Object B (iii) Object C (iv) Object D - Fig6.37 Position-Time - Object A straight up constant velocity, B horizontal rest, C curved increasing slope accelerating, D straight down constant velocity.
Ans: Only Object C - curved shows acceleration net needed Fig6.15 Fig6.16
Q7. A sailor jumps out from a small boat to the shore (Fig6.38). As sailor jumps forward, will boat move? If yes, in which direction and why. Fig6.38 A sailor jumping forward.
Ans: Yes boat moves backward third law sailor pushes boat backward boat pushes sailor forward Fig6.23
Q8. During high jump event landing mat or sand bed placed for athlete to fall upon Fig6.39 Explain reason. Fig6.39 Landing mat for high jump.
Ans: Time increased acceleration decreased force decreased injury minimized cricket catch Fig6.18 airbag Fig6.19
फ्रेम 2/4 - अग्निवीर - Q9 से Q13 पूरे पद पेज114 Fig6.40 Fig6.41
Q9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During collision: (i) loaded cart exerts force larger magnitude on empty (ii) empty exerts larger magnitude on loaded (iii) neither exerts force on other (iv) loaded and empty both exert equal magnitude of force on each other.
Ans: (iv) equal magnitude opposite direction third law acceleration different due mass Example6.8
Q10. The acceleration-mass graph for acceleration produced by a force on objects of different masses is plotted in Fig6.40. Plot force-mass graph for this case. Fig6.40 Acceleration vs Mass hyperbola 2.5,5,7.5,10 kg.
Ans: F=ma constant 10 N - force-mass horizontal straight line 10 N Eq6.2
Q11. The velocity-time graph of object mass 10 kg moving along straight line shown in Fig6.41 Calculate force acting on object by using graph. Fig6.41 Velocity Time 0-4s 10-20, 4-8s 20-30, Time 0-8 Velocity 0-30.
Ans: Slope a=(30-10)/8=2.5 F=ma=10×2.5=25 N Example6.6 Fig6.21
Q12. A bullet mass 50 g moving speed 100 m s⁻¹ enters heavy stationary wooden block stops after penetrating distance 50 cm Estimate stopping force acting on bullet assume bullet undergoes constant acceleration within block.
Ans: m0.05 u100 v0 s0.5 v²=u²+2as a=-10000 F=ma=500 N opposite
Q13. An ace footballer converted penalty shot by kicking football with speed 108 km h⁻¹ Estimated force they imparted was 800 N Mass football 0.4 kg Calculate time of contact between foot and ball.
Ans: 108km/h=30m/s Ft=mv t=0.4×30/800=0.015 s momentum=mv rate change = net force
Q9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During collision: (i) loaded cart exerts force larger magnitude on empty (ii) empty exerts larger magnitude on loaded (iii) neither exerts force on other (iv) loaded and empty both exert equal magnitude of force on each other.
Ans: (iv) equal magnitude opposite direction third law acceleration different due mass Example6.8
Q10. The acceleration-mass graph for acceleration produced by a force on objects of different masses is plotted in Fig6.40. Plot force-mass graph for this case. Fig6.40 Acceleration vs Mass hyperbola 2.5,5,7.5,10 kg.
Ans: F=ma constant 10 N - force-mass horizontal straight line 10 N Eq6.2
Q11. The velocity-time graph of object mass 10 kg moving along straight line shown in Fig6.41 Calculate force acting on object by using graph. Fig6.41 Velocity Time 0-4s 10-20, 4-8s 20-30, Time 0-8 Velocity 0-30.
Ans: Slope a=(30-10)/8=2.5 F=ma=10×2.5=25 N Example6.6 Fig6.21
Q12. A bullet mass 50 g moving speed 100 m s⁻¹ enters heavy stationary wooden block stops after penetrating distance 50 cm Estimate stopping force acting on bullet assume bullet undergoes constant acceleration within block.
Ans: m0.05 u100 v0 s0.5 v²=u²+2as a=-10000 F=ma=500 N opposite
Q13. An ace footballer converted penalty shot by kicking football with speed 108 km h⁻¹ Estimated force they imparted was 800 N Mass football 0.4 kg Calculate time of contact between foot and ball.
Ans: 108km/h=30m/s Ft=mv t=0.4×30/800=0.015 s momentum=mv rate change = net force
फ्रेम 3/4 - अग्निवीर - Q14 से Q16 पूरे पद पेज114-115 Fig6.42 Fig6.43 और MCQ 25 ट्रिक
Q14. An object mass 2 kg moving constant velocity 10 m s⁻¹ encounters rough patch where force friction 7 N At same time additional constant force 3 N opposing motion applied After entering rough patch how much distance object travels before coming to rest?
Ans: Friction 7+3=10 opposite a=10/2=5 opposite u10 v0 v²=u²-2as s=10m
Q15. A tractor pulls harrow mass m1 with net force F resulting acceleration a1 Same tractor pulls trolley mass m2 with force F producing acceleration a2 If tractor now pulls trolley with harrow placed on it same force F then obtain expression for resulting acceleration in terms a1 and a2 Ignore friction.
Ans: F=m1a1 F=m2a2 total m1+m2 F=(m1+m2)a a=F/(m1+m2) 1/a=1/a1+1/a2 a=a1a2/(a1+a2) Eq6.4 Fig6.34
Q16. When pole of bar magnet brought close to magnetic compass bar magnet and compass needle which is also magnet exert magnetic force on each other As per Newton third law both forces equal magnitude opposite direction However compass needle moves whereas bar magnet does not move Fig6.42 Explain why. Fig6.42 Bar magnet and magnetic compass.
Ans: Forces equal magnitude opposite but masses different a=F/m different needle light inertia small moves easily bar heavy inertia large Example6.7 Earth fruit Fig6.33 Example6.8 bullet gun Note page110 Even though forces equal magnitudes they do not in general produce equal acceleration because masses different.
In real world friction always present whenever surfaces contact Scientists trying find ways reduce friction using lubricants coatings texturing streamlining shapes magnetic levitation p115.
Fig6.43 Rope around rough tree branch post heavy bucket one turn hold other end now add one more turn repeat each extra turn increases grip between rope and branch increasing friction reducing force required making much easier hold same load reduction much larger than expected small changes contact large changes force same way friction rope post allows large ships held safely at pier.
Newton laws describe motion enormous range scales everyday objects planets stars Need modification only very close massive objects extremely high speeds near speed light very small atomic scales.
Q14. An object mass 2 kg moving constant velocity 10 m s⁻¹ encounters rough patch where force friction 7 N At same time additional constant force 3 N opposing motion applied After entering rough patch how much distance object travels before coming to rest?
Ans: Friction 7+3=10 opposite a=10/2=5 opposite u10 v0 v²=u²-2as s=10m
Q15. A tractor pulls harrow mass m1 with net force F resulting acceleration a1 Same tractor pulls trolley mass m2 with force F producing acceleration a2 If tractor now pulls trolley with harrow placed on it same force F then obtain expression for resulting acceleration in terms a1 and a2 Ignore friction.
Ans: F=m1a1 F=m2a2 total m1+m2 F=(m1+m2)a a=F/(m1+m2) 1/a=1/a1+1/a2 a=a1a2/(a1+a2) Eq6.4 Fig6.34
Q16. When pole of bar magnet brought close to magnetic compass bar magnet and compass needle which is also magnet exert magnetic force on each other As per Newton third law both forces equal magnitude opposite direction However compass needle moves whereas bar magnet does not move Fig6.42 Explain why. Fig6.42 Bar magnet and magnetic compass.
Ans: Forces equal magnitude opposite but masses different a=F/m different needle light inertia small moves easily bar heavy inertia large Example6.7 Earth fruit Fig6.33 Example6.8 bullet gun Note page110 Even though forces equal magnitudes they do not in general produce equal acceleration because masses different.
In real world friction always present whenever surfaces contact Scientists trying find ways reduce friction using lubricants coatings texturing streamlining shapes magnetic levitation p115.
Fig6.43 Rope around rough tree branch post heavy bucket one turn hold other end now add one more turn repeat each extra turn increases grip between rope and branch increasing friction reducing force required making much easier hold same load reduction much larger than expected small changes contact large changes force same way friction rope post allows large ships held safely at pier.
Newton laws describe motion enormous range scales everyday objects planets stars Need modification only very close massive objects extremely high speeds near speed light very small atomic scales.
फ्रेम 4/4 - अग्निवीर - MCQ 25 ट्रिक पक्का 5 प्रश्न पूरे पद
MCQ 25 पूरे पद:
1. SI unit force newton N not small n page95
2. Net force 10N 6N 16N right 4N right 4N left Example6.1 page96 Fig6.6
3. Distance larger friction less Activity6.1 wooden cemented laminated polished marble page98-99 Fig6.12
4. First law rest remains rest constant velocity remains unless net force acts inertia Galileo indefinite page100
5. Net zero two possibilities rest or constant velocity Fig6.15 Fig6.16 Example6.2 Example6.3 page101
6. Second law definition a=F/m Eq6.1 F=ma Eq6.2 F=mg Eq6.3 g9.8~10 momentum mv rate change proportional net force page102-104 Activity6.3 Fig6.17
7. Cricket catch hands backwards time increased acceleration decreased force decreased injury minimized Fig6.18 page104-105
8. Airbag inflated soft cushion time increased acceleration decreased force decreased Fig6.19 page105
9. Coconut high velocity short time large force breaks shell Fig6.20 page105
10. Weightlifter 10kg each side bar10kg total30kg F=mg=294N upward page105 Example6.4
11. Block 25kg friction50N (i)50N net0 stationary (ii)55N net5N a0.2 s0.4m Example6.5 page105
12. Sports car 1500kg velocity-time Fig6.21 0-5s a2 F3000N east 5-10s 0 10-15s a-2 F-3000N west Example6.6 page106
13. Toy car 100g 0.5m/s constant velocity net0 page106 Q6
14. Two children different masses swings identical acceleration larger mass larger force F=ma page106 Q7
15. Glass bubble wrap hay time increased force decreased page106 Q8
16. Walking foot ground backward ground forward friction helps Fig6.25 page107
17. Spring balance same reading forces equal opposite Fig6.26 page108
18. Tree climbing foot down friction up Fig6.27 page108
19. Canoe water backward paddle forward equal opposite different objects Fig6.28 page108 Think It Over
20. Balloon straw thread Fig6.29 air out opposite balloon, rocket gas down force up Fig6.30 Vikram lander Chandrayaan-3 page109
21. Earth fruit equal opposite gravitational Fig6.33 page110 Example6.7 acceleration Earth extremely small F/m mass large
22. Bullet 0.1kg gun5kg force2N gun0.4 bullet20 acceleration not equal masses different Example6.8 page110
23. System two boxes m1 m2 string tension T internal F external a=F/(m1+m2) Eq6.4 Fig6.34 Fig6.35 page111
24. Exercise Q1 to Q16 full page112-114 as above
25. Rope tree branch one more turn grip increases friction reduces force required Fig6.43 page115
MCQ 25 पूरे पद:
1. SI unit force newton N not small n page95
2. Net force 10N 6N 16N right 4N right 4N left Example6.1 page96 Fig6.6
3. Distance larger friction less Activity6.1 wooden cemented laminated polished marble page98-99 Fig6.12
4. First law rest remains rest constant velocity remains unless net force acts inertia Galileo indefinite page100
5. Net zero two possibilities rest or constant velocity Fig6.15 Fig6.16 Example6.2 Example6.3 page101
6. Second law definition a=F/m Eq6.1 F=ma Eq6.2 F=mg Eq6.3 g9.8~10 momentum mv rate change proportional net force page102-104 Activity6.3 Fig6.17
7. Cricket catch hands backwards time increased acceleration decreased force decreased injury minimized Fig6.18 page104-105
8. Airbag inflated soft cushion time increased acceleration decreased force decreased Fig6.19 page105
9. Coconut high velocity short time large force breaks shell Fig6.20 page105
10. Weightlifter 10kg each side bar10kg total30kg F=mg=294N upward page105 Example6.4
11. Block 25kg friction50N (i)50N net0 stationary (ii)55N net5N a0.2 s0.4m Example6.5 page105
12. Sports car 1500kg velocity-time Fig6.21 0-5s a2 F3000N east 5-10s 0 10-15s a-2 F-3000N west Example6.6 page106
13. Toy car 100g 0.5m/s constant velocity net0 page106 Q6
14. Two children different masses swings identical acceleration larger mass larger force F=ma page106 Q7
15. Glass bubble wrap hay time increased force decreased page106 Q8
16. Walking foot ground backward ground forward friction helps Fig6.25 page107
17. Spring balance same reading forces equal opposite Fig6.26 page108
18. Tree climbing foot down friction up Fig6.27 page108
19. Canoe water backward paddle forward equal opposite different objects Fig6.28 page108 Think It Over
20. Balloon straw thread Fig6.29 air out opposite balloon, rocket gas down force up Fig6.30 Vikram lander Chandrayaan-3 page109
21. Earth fruit equal opposite gravitational Fig6.33 page110 Example6.7 acceleration Earth extremely small F/m mass large
22. Bullet 0.1kg gun5kg force2N gun0.4 bullet20 acceleration not equal masses different Example6.8 page110
23. System two boxes m1 m2 string tension T internal F external a=F/(m1+m2) Eq6.4 Fig6.34 Fig6.35 page111
24. Exercise Q1 to Q16 full page112-114 as above
25. Rope tree branch one more turn grip increases friction reduces force required Fig6.43 page115
✅ भाग 4 - अग्निवीर - MCQ 25 ट्रिक - पक्का 5 प्रश्न अग्निवीर में इसी से - दांग से शिक्षा - शिक्षा का उन्मूलन नहीं ✅
भाग 1 सामान्य परिचय, भाग 2 परिभाषा सिद्धांत सूत्र, भाग 3 अभ्यास, भाग 4 अग्निवीर
भाग 1 सामान्य परिचय, भाग 2 परिभाषा सिद्धांत सूत्र, भाग 3 अभ्यास, भाग 4 अग्निवीर
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