Page 1 14. Let all the particles meet at time t (seconds) \ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t t :2 = 5 4 : Option (c) is correct. JEE Corner Assertion and Reason 1. For stopping car : Maximum retardation = Maximum frictional force m = = = m m m N m mg m g v u as 2 2 2 = + 0 2 2 2 = + - v g d ( )( ) m [ Q Initial velocity = v] \ d v g = 2 2m For circular turn of car : Centripetal force = Maximum frictional force = m mg \ mv d mg 2 ¢ = m i.e., safe radius = ¢ d = v g 2 m Þ d d ¢ = 2 Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. 2. a v v ® ® ® = - av B A AB t = ® BD t AB i.e., | | | | a BD ® = ® = av t v t AB AB 2 Now, | | | | v AB ® = ® = av t R t AB AB 2 \ | | | | a v ® ® = = av av v R 2 2 w (angular velocity) Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. 4. If speed is constant angle between v ® and a ® will always be 90° \ v a ® ® × = 0 150 | Mechanics-1 B D A v B v A ® ® v A ® â€“1 2.5 ms B C A â€“1 1 ms â€“1 2 ms a = a c w q ® ® ® Page 2 14. Let all the particles meet at time t (seconds) \ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t t :2 = 5 4 : Option (c) is correct. JEE Corner Assertion and Reason 1. For stopping car : Maximum retardation = Maximum frictional force m = = = m m m N m mg m g v u as 2 2 2 = + 0 2 2 2 = + - v g d ( )( ) m [ Q Initial velocity = v] \ d v g = 2 2m For circular turn of car : Centripetal force = Maximum frictional force = m mg \ mv d mg 2 ¢ = m i.e., safe radius = ¢ d = v g 2 m Þ d d ¢ = 2 Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. 2. a v v ® ® ® = - av B A AB t = ® BD t AB i.e., | | | | a BD ® = ® = av t v t AB AB 2 Now, | | | | v AB ® = ® = av t R t AB AB 2 \ | | | | a v ® ® = = av av v R 2 2 w (angular velocity) Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. 4. If speed is constant angle between v ® and a ® will always be 90° \ v a ® ® × = 0 150 | Mechanics-1 B D A v B v A ® ® v A ® â€“1 2.5 ms B C A â€“1 1 ms â€“1 2 ms a = a c w q ® ® ® If speed is increasing angle q between v ® and a ® will be less than 90°. \ v a ® ® × will be positive. If speed is decreasing angle between v ® and a ® will be greater than 90°. \ v a ® ® × will be negative. Assertion is correct. Rea son w ® ® × = v 0 as both are perpendicular to each other. Reason is also true but not the correct explanation of the assertion. Option (b) is correct. 5. v i ® = 2 ^ ms -1 a i j ® = - + ^ ^ 2 ms -2 \ a j ® - = c 2 2 ^ ms a a c c ® ® = = | | 2 ms -2 v = = ® | | v 2 ms -1 a v R c = 2 Þ R v a c = 2 = = ( ) 2 2 2 2 m a i T ® = - ^ ms -2 Speed is decreasing (as a T ® is -ive) at a rate of 1 ms -1 per second i.e., 1 ms -2 . Both Assertion and the Reason are correct but reason has nothing to do with the assertion. Option (b) is correct. 6. a a a ® ® ® = + T c (Reason) \ | | a ® = + v r g 2 2 Þ | | a ® > g (Assertion) We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion. Option (a) is correct. 7. At points A and C : The bob is momentarily at rest. i.e., v = 0 (Reason) \ | | a c v R ® = = 2 0 but net acceleration is not zero (see figure) i.e., Assertion is false. \ Option (d) is correct. Circular Motion | 151 a c q v a T a ® ® ® ® R 2j v = 2z a net i x â€“ ® a c q v a T a ® ® ® ® a = g T a c a A ® ® ® B g g A C ® ® Page 3 14. Let all the particles meet at time t (seconds) \ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t t :2 = 5 4 : Option (c) is correct. JEE Corner Assertion and Reason 1. For stopping car : Maximum retardation = Maximum frictional force m = = = m m m N m mg m g v u as 2 2 2 = + 0 2 2 2 = + - v g d ( )( ) m [ Q Initial velocity = v] \ d v g = 2 2m For circular turn of car : Centripetal force = Maximum frictional force = m mg \ mv d mg 2 ¢ = m i.e., safe radius = ¢ d = v g 2 m Þ d d ¢ = 2 Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. 2. a v v ® ® ® = - av B A AB t = ® BD t AB i.e., | | | | a BD ® = ® = av t v t AB AB 2 Now, | | | | v AB ® = ® = av t R t AB AB 2 \ | | | | a v ® ® = = av av v R 2 2 w (angular velocity) Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. 4. If speed is constant angle between v ® and a ® will always be 90° \ v a ® ® × = 0 150 | Mechanics-1 B D A v B v A ® ® v A ® â€“1 2.5 ms B C A â€“1 1 ms â€“1 2 ms a = a c w q ® ® ® If speed is increasing angle q between v ® and a ® will be less than 90°. \ v a ® ® × will be positive. If speed is decreasing angle between v ® and a ® will be greater than 90°. \ v a ® ® × will be negative. Assertion is correct. Rea son w ® ® × = v 0 as both are perpendicular to each other. Reason is also true but not the correct explanation of the assertion. Option (b) is correct. 5. v i ® = 2 ^ ms -1 a i j ® = - + ^ ^ 2 ms -2 \ a j ® - = c 2 2 ^ ms a a c c ® ® = = | | 2 ms -2 v = = ® | | v 2 ms -1 a v R c = 2 Þ R v a c = 2 = = ( ) 2 2 2 2 m a i T ® = - ^ ms -2 Speed is decreasing (as a T ® is -ive) at a rate of 1 ms -1 per second i.e., 1 ms -2 . Both Assertion and the Reason are correct but reason has nothing to do with the assertion. Option (b) is correct. 6. a a a ® ® ® = + T c (Reason) \ | | a ® = + v r g 2 2 Þ | | a ® > g (Assertion) We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion. Option (a) is correct. 7. At points A and C : The bob is momentarily at rest. i.e., v = 0 (Reason) \ | | a c v R ® = = 2 0 but net acceleration is not zero (see figure) i.e., Assertion is false. \ Option (d) is correct. Circular Motion | 151 a c q v a T a ® ® ® ® R 2j v = 2z a net i x â€“ ® a c q v a T a ® ® ® ® a = g T a c a A ® ® ® B g g A C ® ® 8. v (speed) = - 4 12 t For t < 3 (time unit) speed is negative, which canâ€™t describe a motion. Thus, assertion is correct. As speed can be changed linearly with time, the reason is false. Option (c) is correct. 9. In circular motion the acceleration changes regularly where as in projectile motion it is constant. Thus, in circular motion we can't apply v u a ® ® ® = + t directly, whereas in projectile motion we can say reason that in circular motion gravity has no role is wrong. Option (c) is correct. 10. N mg = cos q Therefore, assertion is wrong. Particle performs circular motion due to Nsinq \ mv r N 2 = sinq Ncosq is balanced by mg (weight of particle). Acceleration is not along the surface of the funnel. It is along the centre O of the circle. Thus, reason is true. Option (d) is correct. 11. Centripetal force mv r N mg 2 æ è ç ö ø ÷ = + i.e., Centripetal force (reason ) ³ wt mg ( ) of water for N = 0, v gr = If at the top of the circular path v gr ³ i.e., if bucket moved fast, the water will not fall (Assertion). As assertion and reason both are true and reason is the correct explanation of the assertion of the option would be (a). Objective Questions (Level 2) Single Correct Option 1. KE 1 2 2 mv æ è ç ö ø ÷ = Change in PE 1 2 2 k x ( ) D é ë ê ù û ú Dx = Length of spring (Collar at B) - Length of spring (Collar at A) = + + ( ) 7 5 5 2 2 m - 7m = 6 m Thus, 1 2 2 1 2 200 6 2 2 ´ ´ = ´ ´ v i.e., v 2 3600 = \ Normal reaction = mv r 2 = ´ 2 3600 5 = 1440 N Option (a) is correct. 2. The particle will remain in equilibrium till w is constant. Any change in the value of w will displace the particle up (if w increase) and down (if w decreases). Thus, the equilibrium is unstable. Option (b) is correct. 3. Centripetal force = m mg or mr mg w m 2 = or 5 4 2 a g w m = or w 2 4 15 = g a asm = æ è ç ö ø ÷ 1 3 Option (d) is correct. 152 | Mechanics-1 mg cos q mg N O N sinq q N mg Page 4 14. Let all the particles meet at time t (seconds) \ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t t :2 = 5 4 : Option (c) is correct. JEE Corner Assertion and Reason 1. For stopping car : Maximum retardation = Maximum frictional force m = = = m m m N m mg m g v u as 2 2 2 = + 0 2 2 2 = + - v g d ( )( ) m [ Q Initial velocity = v] \ d v g = 2 2m For circular turn of car : Centripetal force = Maximum frictional force = m mg \ mv d mg 2 ¢ = m i.e., safe radius = ¢ d = v g 2 m Þ d d ¢ = 2 Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. 2. a v v ® ® ® = - av B A AB t = ® BD t AB i.e., | | | | a BD ® = ® = av t v t AB AB 2 Now, | | | | v AB ® = ® = av t R t AB AB 2 \ | | | | a v ® ® = = av av v R 2 2 w (angular velocity) Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. 4. If speed is constant angle between v ® and a ® will always be 90° \ v a ® ® × = 0 150 | Mechanics-1 B D A v B v A ® ® v A ® â€“1 2.5 ms B C A â€“1 1 ms â€“1 2 ms a = a c w q ® ® ® If speed is increasing angle q between v ® and a ® will be less than 90°. \ v a ® ® × will be positive. If speed is decreasing angle between v ® and a ® will be greater than 90°. \ v a ® ® × will be negative. Assertion is correct. Rea son w ® ® × = v 0 as both are perpendicular to each other. Reason is also true but not the correct explanation of the assertion. Option (b) is correct. 5. v i ® = 2 ^ ms -1 a i j ® = - + ^ ^ 2 ms -2 \ a j ® - = c 2 2 ^ ms a a c c ® ® = = | | 2 ms -2 v = = ® | | v 2 ms -1 a v R c = 2 Þ R v a c = 2 = = ( ) 2 2 2 2 m a i T ® = - ^ ms -2 Speed is decreasing (as a T ® is -ive) at a rate of 1 ms -1 per second i.e., 1 ms -2 . Both Assertion and the Reason are correct but reason has nothing to do with the assertion. Option (b) is correct. 6. a a a ® ® ® = + T c (Reason) \ | | a ® = + v r g 2 2 Þ | | a ® > g (Assertion) We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion. Option (a) is correct. 7. At points A and C : The bob is momentarily at rest. i.e., v = 0 (Reason) \ | | a c v R ® = = 2 0 but net acceleration is not zero (see figure) i.e., Assertion is false. \ Option (d) is correct. Circular Motion | 151 a c q v a T a ® ® ® ® R 2j v = 2z a net i x â€“ ® a c q v a T a ® ® ® ® a = g T a c a A ® ® ® B g g A C ® ® 8. v (speed) = - 4 12 t For t < 3 (time unit) speed is negative, which canâ€™t describe a motion. Thus, assertion is correct. As speed can be changed linearly with time, the reason is false. Option (c) is correct. 9. In circular motion the acceleration changes regularly where as in projectile motion it is constant. Thus, in circular motion we can't apply v u a ® ® ® = + t directly, whereas in projectile motion we can say reason that in circular motion gravity has no role is wrong. Option (c) is correct. 10. N mg = cos q Therefore, assertion is wrong. Particle performs circular motion due to Nsinq \ mv r N 2 = sinq Ncosq is balanced by mg (weight of particle). Acceleration is not along the surface of the funnel. It is along the centre O of the circle. Thus, reason is true. Option (d) is correct. 11. Centripetal force mv r N mg 2 æ è ç ö ø ÷ = + i.e., Centripetal force (reason ) ³ wt mg ( ) of water for N = 0, v gr = If at the top of the circular path v gr ³ i.e., if bucket moved fast, the water will not fall (Assertion). As assertion and reason both are true and reason is the correct explanation of the assertion of the option would be (a). Objective Questions (Level 2) Single Correct Option 1. KE 1 2 2 mv æ è ç ö ø ÷ = Change in PE 1 2 2 k x ( ) D é ë ê ù û ú Dx = Length of spring (Collar at B) - Length of spring (Collar at A) = + + ( ) 7 5 5 2 2 m - 7m = 6 m Thus, 1 2 2 1 2 200 6 2 2 ´ ´ = ´ ´ v i.e., v 2 3600 = \ Normal reaction = mv r 2 = ´ 2 3600 5 = 1440 N Option (a) is correct. 2. The particle will remain in equilibrium till w is constant. Any change in the value of w will displace the particle up (if w increase) and down (if w decreases). Thus, the equilibrium is unstable. Option (b) is correct. 3. Centripetal force = m mg or mr mg w m 2 = or 5 4 2 a g w m = or w 2 4 15 = g a asm = æ è ç ö ø ÷ 1 3 Option (d) is correct. 152 | Mechanics-1 mg cos q mg N O N sinq q N mg 4. Acceleration at B = Acceleration at A \ v r g 2 = sinq or 2 1 gr r g ( cos ) sin - = q q or 2 1 ( cos ) sin - = q q or [ ( cos )] cos 2 1 1 2 2 - = - q q or ( cos ) (cos ) 5 3 1 0 q q - - = As cos q = 1, i.e., q = ° 0 is not possible. cos q = 3 5 i.e., q = - cos 1 3 5 Option (c) is correct. 5. At point P (for the circular motion) mg N mv R cos q - = 2 If at point P skier leaves the hemisphere. N =0 \ mg mv R cosq = 2 or mg m R g h R cosq = + æ è ç ö ø ÷ 2 4 or mg m R g R R cos ( cos ) q q = - + é ë ê ù û ú 2 1 4 Þ cos ( cos ) q q = - + é ë ê ù û ú 2 1 1 4 i.e., q = - cos 1 5 6 Option (c) is correct. 6. Velocity at B v gh = 2 = - 2 2 1 gR(cos cos ) q q = ° - ° 2 37 53 gR(cos cos ) = - æ è ç ö ø ÷ 2 4 3 3 5 gR = 2 5 gR For the circular motion at B, when block just leaves the track v R g 2 1 2 = cos q or R v g R 1 2 2 2 2 5 = = cos cos q q = × ° 2 5 37 R cos = × æ è ç ö ø ÷ 2 5 4 5 R = R 2 Option (c) is correct. 7. mg mv a cosq= 2 or g g a a cosq = + × × 4 2 4 2 or g g a a u g a a - = + × × 4 2 4 2 or g a u g a × = + 3 4 2 4 2 Þ u ag = 2 Option (c) is correct. 8. v r r 2 2 4 = Þ v r = 2 \ Momentum = mv = 2m r 9. N mr cosq w = 2 and N mg sin q = \ N m g r = + 2 2 4 w = + æ è ç ö ø ÷ m g r T 2 2 4 2p = + æ è ç ö ø ÷ 10 10 2 2 2 4 ( ) 0.5 1.5 p a =128 N Option (b) is correct. 10. q w a = + 0 2 1 2 t t = + é ë ê ù û ú w a 0 1 2 t t Circular Motion | 153 Page 5 14. Let all the particles meet at time t (seconds) \ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t t :2 = 5 4 : Option (c) is correct. JEE Corner Assertion and Reason 1. For stopping car : Maximum retardation = Maximum frictional force m = = = m m m N m mg m g v u as 2 2 2 = + 0 2 2 2 = + - v g d ( )( ) m [ Q Initial velocity = v] \ d v g = 2 2m For circular turn of car : Centripetal force = Maximum frictional force = m mg \ mv d mg 2 ¢ = m i.e., safe radius = ¢ d = v g 2 m Þ d d ¢ = 2 Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. 2. a v v ® ® ® = - av B A AB t = ® BD t AB i.e., | | | | a BD ® = ® = av t v t AB AB 2 Now, | | | | v AB ® = ® = av t R t AB AB 2 \ | | | | a v ® ® = = av av v R 2 2 w (angular velocity) Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. 4. If speed is constant angle between v ® and a ® will always be 90° \ v a ® ® × = 0 150 | Mechanics-1 B D A v B v A ® ® v A ® â€“1 2.5 ms B C A â€“1 1 ms â€“1 2 ms a = a c w q ® ® ® If speed is increasing angle q between v ® and a ® will be less than 90°. \ v a ® ® × will be positive. If speed is decreasing angle between v ® and a ® will be greater than 90°. \ v a ® ® × will be negative. Assertion is correct. Rea son w ® ® × = v 0 as both are perpendicular to each other. Reason is also true but not the correct explanation of the assertion. Option (b) is correct. 5. v i ® = 2 ^ ms -1 a i j ® = - + ^ ^ 2 ms -2 \ a j ® - = c 2 2 ^ ms a a c c ® ® = = | | 2 ms -2 v = = ® | | v 2 ms -1 a v R c = 2 Þ R v a c = 2 = = ( ) 2 2 2 2 m a i T ® = - ^ ms -2 Speed is decreasing (as a T ® is -ive) at a rate of 1 ms -1 per second i.e., 1 ms -2 . Both Assertion and the Reason are correct but reason has nothing to do with the assertion. Option (b) is correct. 6. a a a ® ® ® = + T c (Reason) \ | | a ® = + v r g 2 2 Þ | | a ® > g (Assertion) We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion. Option (a) is correct. 7. At points A and C : The bob is momentarily at rest. i.e., v = 0 (Reason) \ | | a c v R ® = = 2 0 but net acceleration is not zero (see figure) i.e., Assertion is false. \ Option (d) is correct. Circular Motion | 151 a c q v a T a ® ® ® ® R 2j v = 2z a net i x â€“ ® a c q v a T a ® ® ® ® a = g T a c a A ® ® ® B g g A C ® ® 8. v (speed) = - 4 12 t For t < 3 (time unit) speed is negative, which canâ€™t describe a motion. Thus, assertion is correct. As speed can be changed linearly with time, the reason is false. Option (c) is correct. 9. In circular motion the acceleration changes regularly where as in projectile motion it is constant. Thus, in circular motion we can't apply v u a ® ® ® = + t directly, whereas in projectile motion we can say reason that in circular motion gravity has no role is wrong. Option (c) is correct. 10. N mg = cos q Therefore, assertion is wrong. Particle performs circular motion due to Nsinq \ mv r N 2 = sinq Ncosq is balanced by mg (weight of particle). Acceleration is not along the surface of the funnel. It is along the centre O of the circle. Thus, reason is true. Option (d) is correct. 11. Centripetal force mv r N mg 2 æ è ç ö ø ÷ = + i.e., Centripetal force (reason ) ³ wt mg ( ) of water for N = 0, v gr = If at the top of the circular path v gr ³ i.e., if bucket moved fast, the water will not fall (Assertion). As assertion and reason both are true and reason is the correct explanation of the assertion of the option would be (a). Objective Questions (Level 2) Single Correct Option 1. KE 1 2 2 mv æ è ç ö ø ÷ = Change in PE 1 2 2 k x ( ) D é ë ê ù û ú Dx = Length of spring (Collar at B) - Length of spring (Collar at A) = + + ( ) 7 5 5 2 2 m - 7m = 6 m Thus, 1 2 2 1 2 200 6 2 2 ´ ´ = ´ ´ v i.e., v 2 3600 = \ Normal reaction = mv r 2 = ´ 2 3600 5 = 1440 N Option (a) is correct. 2. The particle will remain in equilibrium till w is constant. Any change in the value of w will displace the particle up (if w increase) and down (if w decreases). Thus, the equilibrium is unstable. Option (b) is correct. 3. Centripetal force = m mg or mr mg w m 2 = or 5 4 2 a g w m = or w 2 4 15 = g a asm = æ è ç ö ø ÷ 1 3 Option (d) is correct. 152 | Mechanics-1 mg cos q mg N O N sinq q N mg 4. Acceleration at B = Acceleration at A \ v r g 2 = sinq or 2 1 gr r g ( cos ) sin - = q q or 2 1 ( cos ) sin - = q q or [ ( cos )] cos 2 1 1 2 2 - = - q q or ( cos ) (cos ) 5 3 1 0 q q - - = As cos q = 1, i.e., q = ° 0 is not possible. cos q = 3 5 i.e., q = - cos 1 3 5 Option (c) is correct. 5. At point P (for the circular motion) mg N mv R cos q - = 2 If at point P skier leaves the hemisphere. N =0 \ mg mv R cosq = 2 or mg m R g h R cosq = + æ è ç ö ø ÷ 2 4 or mg m R g R R cos ( cos ) q q = - + é ë ê ù û ú 2 1 4 Þ cos ( cos ) q q = - + é ë ê ù û ú 2 1 1 4 i.e., q = - cos 1 5 6 Option (c) is correct. 6. Velocity at B v gh = 2 = - 2 2 1 gR(cos cos ) q q = ° - ° 2 37 53 gR(cos cos ) = - æ è ç ö ø ÷ 2 4 3 3 5 gR = 2 5 gR For the circular motion at B, when block just leaves the track v R g 2 1 2 = cos q or R v g R 1 2 2 2 2 5 = = cos cos q q = × ° 2 5 37 R cos = × æ è ç ö ø ÷ 2 5 4 5 R = R 2 Option (c) is correct. 7. mg mv a cosq= 2 or g g a a cosq = + × × 4 2 4 2 or g g a a u g a a - = + × × 4 2 4 2 or g a u g a × = + 3 4 2 4 2 Þ u ag = 2 Option (c) is correct. 8. v r r 2 2 4 = Þ v r = 2 \ Momentum = mv = 2m r 9. N mr cosq w = 2 and N mg sin q = \ N m g r = + 2 2 4 w = + æ è ç ö ø ÷ m g r T 2 2 4 2p = + æ è ç ö ø ÷ 10 10 2 2 2 4 ( ) 0.5 1.5 p a =128 N Option (b) is correct. 10. q w a = + 0 2 1 2 t t = + é ë ê ù û ú w a 0 1 2 t t Circular Motion | 153 = + - × æ è ç ö ø ÷ æ è ç ö ø ÷ é ë ê ê ù û ú ú v R R v R R v t 1 2 1 4 4 2 p p = - é ë ê ù û ú v R v R R v 2 4p = 4p =2 rev. 11. H gT = 1 2 2 Þ T H g = 2 w p p = = 2 2 2 T g H = p 2g H 12. w (minute hand) = - 2 3600 1 p rad s w (second hand) = 2 60 p rads -1 For second hand to meet minute hand for the first time. 2p + Angle moved by minute hand in t second = Angle moved by second hand in t second or 2 2 3600 2 60 p p p + = t t 1 60 3600 = - t t 1 60 59 60 = ´ t Þ t = 3600 59 s Option (d) is correct. 13. ( ) ( ) ( ) PR QR PQ 2 2 2 + = ( ) ( ) v v A B × + × = 2 2 30 2 2 2 v v A B 2 2 225 + = â€¦(i) Further, PR PQ = ° cos30 v A × = 2 30 3 2 Þ v A =7.5 3 ms -1 Substituting value of v A in Eq. (i), v B = 7.5 ms -1 14. mg N mv r cos q - = 2 When breaks off N = 0 \ mg mv r cos q = 2 or g gr r cos ( cos ) q q = - 2 1 or cos ( cos ) q q = - 2 1 or cosq = 2 3 Acceleration of particle when it leaves sphere = gsinq = g 5 3 Option (b) is correct. 154 | Mechanics-1 H R 60° 30° 30 m B' v B t = 2 s R v B = AD q Acceleration = g sin q mg N O mg sin q q Ö5 2 3Read More

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