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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A horizontal force of `1.2kg` is applied on a `1.5kg` block, which rests on a horizontal surface If the coefficient of friction is `0.3` find the acceleration produced in the block. |
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Answer» Correct Answer - `4.9ms^(-2)` Here, `P = 1.2kg = 1.2 xx 9.8 N` `m = 1.5kg, mu = 0.3, a =`? Force of friction `F = muR = mu mg` `=0.3 xx 1.5 xx 9.8 N = 0.45 xx 9.8 N` `:.` force that produces acceleration `f = P - F = 1.2 xx 9.8 - 0.45 xx 9.8` `= 9.8 (1.2 - 0.45) = 9.8 xx 0.75 N` `a = (f)/(m) = (9.8 xx 0.75)/(1.5) = 4.9 ms^(-2)` . |
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| 2. |
A block of mass m slides down on a wedge of mass M as shown in figure .Let `a_(1)` be the asseleration of the acceleration of and `a_(2)` the the acceleration od block 1`N_(1)` is the normal reaction between block and wedge and `N_(2)` the normal reaction between wedge and gound .frication is absent everwhere .Select the correct alternative(s) A. `N_(2)lt(M+m)g`B. `N_(2)ltM(g cos theta-|a_(1)|sintheta)`C. `N_(1)sin theta=M|a_(1)|`D. `ma_(2)=-Ma_(1)` |
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Answer» Correct Answer - a,b,c |
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| 3. |
A hammer weighing 3 kg moving with a velocity of `10ms^(-1)` strikes against the head fo a spike and drives it into a block of wood. If the hammer comes to rest in `0.025` s, find (i) the impulse and (ii) the average force on the head of the spike. |
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Answer» Correct Answer - (i) `-30Ns` (ii) `-3000N` |
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| 4. |
A man of mass 70 kg stands on a weighing machine in a lift, which is moving (a) upwards with a uniform speed of `10ms^(-1)` (b) downwards with a uniform acceleration of `5 ms^(-2)` (c) upwards with a uniform acceleration of `5 ms^(-2)` What would be the readings on the scale in each case ? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ? |
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Answer» Here m = 70 kg , g `= 10 ms^(m/s^(2)` The weighting machine in each case measueres the reaction R i.e. the apparent weight. (a) When the lift moves upwards with a uniform speed, its acceleration is zero . R = mg = `70 xx 10 `= 700 N (b) When the lift moves downwards with a = 5 `ms^(-2)` R = `m(g - a) = 70( 10 - 5)` = 350 N (c ) When the lift moves upwards with a = `5ms^(-2)` R =` m(g + a) = 70 (10 + 5)` = 1050 N (d) If the lift were to come down freely under gravity , downward acc. a = g `therefore` R = m(g-a) = m(g-g)` = Zero. |
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| 5. |
A body wights 8 g when placed on one pan and 18 g when placed on the other pan of a false balance .If the beam is horizontal when both the pans are empty the weght of the body is 4 xg Find value of x. |
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Answer» Correct Answer - 3 |
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| 6. |
A racing car travels on a track (without banking) ABCDEFA ABC is a circular are of radius ` 2 R.CD` and `FA` are strainght paths of length `R` and `DEF` is a circular are of radius `R=100m` The co- efficient of friction on the road is `mu = 0.1` The maximum speed of ther car is `50ms^(-1)` Find the minimum time for completing one round . |
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Answer» As the track is unbanked the necessary centripetal force is provided by force of friction `(m upsilon^(2))/(r) = F = mu R = mu mg :. upsilon= sqrt(mu r g)` For path ABC : length `= (3)/(4) (2pi2R) = 3 pi R = 3 pi xx 100 = 300 pi m` `upsilon_(1)=sqrt(mu 2 Rg) = sqrt( 0.1 xx 2 xx 100 xx 10) = 14.14 m//s` `:. t_(1) = (300 pi)/(14.14) = 66.6s` For path DEF: length `= (1)/(4) (2pi R) = (pi xx 100)/(2) = 50pi` `upsilon_(2)=sqrt(muRg) = sqrt(0.1 xx 100 xx 10 )= 10 m//s` `t_(2) = (50pi)/(10) = 5 pi sec = 15.7 s` For paths `CD` and `FA` length `=R + R =2 R = 200 m` `t_(3) = (200)/(50) = 4.0 s` `:.` Total time for completing one round `t = t_(1) + t_(2) + t_(3) = 66.6 + 15.7 + 4.0 = 86.3 s`. |
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| 7. |
A constant force acting on a body of mass of 5 kg change its speed from `5ms^(-1) "to" 10ms^(-1)` in 10 s without changing the direction of motion. The force acting on the body isA. 1.5NB. 2NC. 2.5ND. 5N |
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Answer» Correct Answer - C Here, 5kg, u=`5ms^(-1), v=10ms^(-1), t=10s` using v=u+at `a=(v-u)/(t)=((10-5)ms^(-1))/(10s)=0.5ms^(-2)` As, F=ma `therefore F=(5kg)(0.5ms^(-2))=2.5N` |
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| 8. |
Given the magnitude and direction of the force acting on a stone of mass 0.1 kg (a) just after it is dropped from the window of a stationary train (b) just after it is dropped from the window of a train running at a constant velocity of `36 km//hr` (c) just after it is droped from the window of a train accelerating with `1 ms^(-2)` (d) lying on the floor of a trin which is accelerating with `1 ms^(-2)` the stone being at rest relative to the train . Neglect the resistance of air throuhout . |
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Answer» (a) Here ,`m = 0 .1 kg , a = + g = 10 m//s^(2)` Net force `F = ma = 0. 1 xx 10 = 1.0 N` This force acts vertically downwards . (b) When the train is runing at a constant velocity its acc = 0 No force acts on the stone due to this motion Therefore force on the stone F = mg ` = 0.1 xx 10 = 1.0 N` This force also acts vertically downwards (c) When the train is accelerating with `1 ms^(-2)` an additional Force `F = ma = 0.1 xx 1 = 0. 1` N acts on the stone in the horizontal direction But once the stone is dropped from tha train F becomes Zero and the net force on the stone is `F = mg = 0.1 xx 10 = 0.1 N` acting vertically downwards (d) As the stone is lying on the floor of the train its acceleration is same as that of the train `:.` force acting on stone ,`F = ma = 0.1 xx 1 = 0.1 N` This force is along the horizontal direction of motion of the train Note that weight of the stone in this case is being balanced by the normal reaction. |
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| 9. |
When forces `F_1`, `F_2`, `F_3`, are acting on a particle of mass m such that `F_2` and `F_3` are mutually perpendicular, then the particle remains stationary. If the force `F_1` is now removed then the acceleration of the particle isA. (a) `F//m`B. (b) `F_2F_3//mF_1`C. (c) `(F_2-F_3)//m`D. (d) `F_2//m` |
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Answer» Correct Answer - A When `F_1`, `F_2` and `F_3` are acting on a particle then the particle remains stationary. This means that the resultant of `F_1`, `F_2` and `F_3` is zero. When `F_1` is removed, `F_2` and `F_3` will remain. But the resultant of `F_2` and `F_3` should be equal and opposite to `F_1`, .i.e. `|vecF_2+vecF_3|=|vecF_1|` `:.` `a=(|vecF_2+vecF_3|)/(m)impliesa=(F_1)/(m)` |
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| 10. |
If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?A. (a) 1 cmB. (b) 2 cmC. (c) 3 cmD. (d) 4 cm |
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Answer» Correct Answer - A `W=DeltaK=FS` `1/2mv^2-1/2m(v/2)^2=Fxx3` …(i) `1/2m(v/2)^2-0=FxxS` …(ii) On dividing `(1//4)/(3//4)=S//3` `:.` `S=1 cm` |
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| 11. |
If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?A. 1 cmB. 2 cmC. 3 cmD. 4 cm |
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Answer» Correct Answer - A (a) Assuming the resistance force or retardation to be constant. `((v)/(2))^(2)=v^(2)-2as_(1)`…….(i) `0=((v)/(2))^(2)-2as_(2)`………(ii) Solving these two equations, we get, `s_(2)=(s_(1))/(3)=1 cm` |
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| 12. |
What is the angular velocity in rad `s^(-1)` of the hour minute and second hand of a clock ? |
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Answer» Correct Answer - `(pi)/(21600) rad s^(-1) ; (pi)/(1800) rad s^(-1) ; (pi)/(30) rad s^(-1)` . (i) The hour hand of a clock completes one revolution in `12` hours i.e `T = 12 h = 12 xx 60 xx 60 s` `omega = (2pi)/(T) = (2pi)/(60 xx 60) = (pi)/(1800) rad s^(-1)` (ii) For minutes hand of the clock, `T =60 min = 60 xx 60 s` `:. omega = (2pi)/(T) = (2pi)/(60 xx60) = (pi)/(1800) rds^(-1)` (iii) For second s hand of the clock `T = 60 sec` `:. omega = (2pi)/(T) = (2pi)/(60 xx60) = (pi)/(30) rds^(-1)` . |
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| 13. |
The line of action of the resultant of two like parallel forces shifts by one-fourth of the distance between the forces when the two forces are interchanged. The ratio of the two forces is:A. `3:4`B. `1:2`C. `3:5`D. `2:3` |
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Answer» Correct Answer - c Let the distance between two parallel forces `F_(1)` and `F_(2)` be `L` if `r_(1)` is distance of the resultant then equating the moments of the forces `F_(1) xx r_(1) = F_(2) xx r_(2)` or `(r_(1))/(r_(2)) = (F_(2))/(F_(1))` Also `r_(1) = (F_(2) L)/(F_(1) + F_(2))` and `r_(2) = (F_(1)L)/(F_(1) + F_(2))` As `r_(1) = r_(2) + (L)/(4)` `:. (F_(2)L)/(F_(1) + F_(2)) = (F_(1)L)/(F_(1) + F_(2)) + (L)/(4)` `(F_(2))/(F_(1) + F_(2)) = (5F_(1) + F_(2))/(4(F_(1) +F_(2))` or `5 F_(1) + F_(2) = 4 F_(2)` `5 F_(1) = 3 F_(2)` or `(F_(1))/(F_(2)) = (3)/(5)` . |
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| 14. |
A uniform rod of length 3L and mass m is suspended from a horizontal roof by two strings of length L and 2L as shown. Then the tension in the left string of length L is : A. mgB. `(mg)/(4)`C. `(mg)/(2)`D. `(mg)/(3)` |
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Answer» Correct Answer - C |
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| 15. |
An electric bulb suspended from the roof of a railway train by a flexible wire shifts through an angle of `19^(@) 48` when the train goes horizontally round a curved path of `200m` radius Find the speed of the train . |
| Answer» Correct Answer - `26.56ms^(-1)` . | |
| 16. |
A machine gun fires 50-g bullets at a speed of 1000 m/s . The gunner , holding the machine in his hands , can exert an average force of 150 N against the gun . Calculate the maximum number of bullets he can fire one minute. |
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Answer» Correct Answer - 180 |
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| 17. |
A machine gun has a mass of `20kg` It fires `30g` bullets at the rate of 400 bullets/s with a sped of `400m//s` What force must be applied on the gun to keep it in position ? |
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Answer» Correct Answer - `4800N` Here, `M = 20 kg, F = `? ,brgt `m = 30 g = 30 xx 10^(-3) kg, n = 400, upsilon = 400m//s` `F=` rate of change of momentum of bullets `= mn upsilon = 30 xx 10^(-3) xx 400 xx 400 = 4800 N` . |
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| 18. |
A machine gun of mass 10 kg fires 20 g bullets at the rate of 10 bullets per second with a speed of 500 `ms^(-1)`. What force is required to hold the gun in position ? |
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Answer» Correct Answer - 100 N |
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| 19. |
`vecF = a hat i + 3hat j+ 6 hat k` and `vec r = 2hat i-6hat j -12 hat k`. The value of `a` for which the angular momentum is conserved isA. -1B. 2C. zeroD. 1 |
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Answer» Correct Answer - A |
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| 20. |
Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. (a) 1:1B. (b) 1:4C. (c) 1:8D. (d) 1:16 |
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Answer» Correct Answer - D `DeltaK=FS` `1/2mu^2=FxxS_1` …(i) `1/2m(4u)_2=FS_2` …(ii) Dividing (i) and (ii), `(u^2)/(16u^2)=(2as_1)/(2as_2)implies1/16=s_1/s_2` |
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| 21. |
What do you think may be the cause of an earthquake ? |
| Answer» There are a series of fractures in earth s crust These are called The San Andreas Fault forces in the earth s interior cause the rocks to slide past each other in a horizontal direction So long as the forces of friction between the two sliding rocks are greater than the forces causing them to slide , the rocks only stretch and twist However , when the forces causing the slide exceed the force of friction the rocks snap back like the release of a stretched rubber band . This movement of rocks with the consequent of tremendous amounts of energy is the earthquake . | |
| 22. |
A ball of mass `1kg` hangs in equilibrium from two strings as shown in If `g = 10m//s^(2)` what are the valuse of tension in strings `OA` and `OB` . |
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Answer» Correct Answer - `5 N ; 5sqrt3N` . As shown in the three force `T_(1), T_(2)` and mg acting at`O` are in equilibrium Applying Lami s theorem we get `(T_(1))/(sin 150^(@)) =(T_(2))/(sin 120^(@)) = (mg)/(sin90^@)` `T_(1) = (mg)/(sin 90^(@)) xxsin 150^(@) = (10 sin 30^(@))/(1) = 5 N` `T_(2) = (mg)/(sin90^(@)) xxsin120^(@) = (10sqrt3//2)/(1) = 5 sqrt3N` . |
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| 23. |
A bullet fired from a gun is more dangerous than an air molecule hitting a person , though both of them have almost the same speed Way ? |
| Answer» Mass of a bullet is much greater than the mass of an air molecule . Therefore , even when both have almost the same speed . Linear momentum of bullet is much larger than the linear momentum of an air molecule Hence the force required to stop a bullet will be much bigger than the force required to stop an air molecule That is way a bullet fired from a gun is more dangerous than an air molecule . | |
| 24. |
The masses `m_(1) m_(2)` and `m_(3)` of the three bodies shown in fig . Are 5 , 2 and 3 kg respectively Calculate the valuse of tension `T_(1) T_(2)` and `T_(3)` when (i) the whole system is going upward with an acceleration of `2 m//s^(2)` (ii) the whole system is stationary ` (g=9.8 m//s^(2))` . . |
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Answer» All the three bodies are moving together with an an upward acc . Of `2 m//s^(2)` The force pulling the system upwards is `T_(1)` and downward pull of gravity is ` (m_(1) + m_(3)) g` According to Newton s 2 nd law of motion `T_(1) - (m_(1) + m_(2) + m_(3) ) g = (m_(1) + m_(3) ) a` or `T_(1) = (m_(1) + m _(2) + m_(3) ) (a + g )` ` = (5+2+3) (2+9.8) = 118 N ` Similarly , for motion of `m_(2)` and `m_(3)` ., we write `T_(2) = (m_(2) + m_(3) ( a+g ) = (2+3) (2 + 9.8)` `= 590 N` and for motion of `m_(3)` `T_(3) = m_(3) (a+g ) = 3 (2+ 9.8 ) = 35 .4 N` (ii) When the whole system is stationary `a= 0 ` Using the same equations as above with a = 0 , ` T_(1) = (m_(1) + m_(2) + m_(3)) g = 10 xx 9.8 = 98 N` `T_(2) = (m_(2) + m_(3)) g = 5 xx 9.8 = 49 N` `T_(3) = m_(3) xx g = 3 xx 9.8 = 29 . 4 N ` |
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| 25. |
What is the tension in a rod of length length L and mass M at a distance y from`F_(1)` when the rod is acted on by two unequal force `F_(1)` and `F_(2) ( lt F_(1))` as shown in. |
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Answer» As is clear from (APC) 3 Net force on the rod ` f = F_(1) - F_(2) ` Acceleration of rod ` a = (f)/(M) = ((F_(1) - F_(2)))/(M) , along F_(1) ` Mass of part AB of the rod ` = ((M)/(L))y ` If T is tension in the rod at B then equation of motion of portion AB of the rod ` F_(1) - T = ma = ((m)/(L) y) ((F_(1) - F_(2))/(M)) ` ` = (F_(1) - F_(2)) (y)/(L)` `T = F_(1) (1- (y)/(L)) + F_(2) ((y)/(L))` ` T = F_(1) (1 - (y)/(L))+ F_(2) (y/(L)) ` This is the desired result |
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| 26. |
Two bodies of masses `11kg` and `11.5kg` are connected by a long light string passing over a smooth pulley Calculate velocity and height ascended/descended by each body at the end of `4s` . |
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Answer» Correct Answer - ` 0.872 m//s, 1.744m` . Here, `m_(1) = 11 kg, m_(2) = 11.5 kg` `:. a = ((m_(2)-m_(1)) g)/(m_(2) + m_(1))= ((11.5 -11))/(11.5 + 11) 9.8 = (4.9)/(22.5)` `=0.218 m//s^(2)` From` upsilon = u + a t = 0 + 0.218 xx4 = 0.872m//s` From ` s = ut + (1)/(2)at^(2) = 0+(1)/(2) xx0.218(4)^(2)` `= 1.744m` . |
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| 27. |
The coefficient of friction between 4 kg and 5kg blocks is 0.2 and between 5kg block and grond is 0.1 respectively Choose the correct statements. A. Minimum force needed to cause system to move is 17NB. When force in 4N static friction at all surfaces in 4N to keep system at rest.C. Maximum acceleration of 4kg block is 2`m//s^(2)`D. Slipping between 4 kg and 5kg blocks start when F is `gt17N` |
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Answer» Correct Answer - c |
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| 28. |
A force ofv `3kg` wt is just sufficient to pull a block of `4kg` over a horizontal surface .What is the angle of friction ? |
| Answer» Correct Answer - `36^(@)52` | |
| 29. |
Figure shows the position-time graph of a particle of mass 4kg. Let the force on the particle for `tlt0` , `0ltt`lt,`4s` , `tgt4s` be `F_(1) , F_(2)` and `F_(3)` respectively. Then A. `F_(1)=F_(2)=F_(3)=0`B. `F_(1)gtF_(2)=F_(3)`C. `F_(1)gtF_(2)gtF_(3)`D. `F_(1)ltF_(2)ltF_(3)` |
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Answer» Correct Answer - A For `t lt 0 and t ge 4s`, the position of the particle is not changing i.e. the particle is at rest. So no force is acting on the particle at these intevals. For `0 lt t lt 4s`. The position of the particle is continuously changing. As the position time graph is a straight line, the motion of the particle is uniform, so acceleration, a=0, Hence no force acts on the particle during the interval also. |
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| 30. |
The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4kg. The magnitude of each impulse is A. `0.8Ns`B. `1.6Ns`C. `0.2Ns`D. `0.4Ns` |
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Answer» Correct Answer - a Here, `m = 0.4kg , ` when ` s = 2m, t = 2 s and u = 0` From `s = ut + (1)/(2)at^(2)` `2 = 0 + (1)/(2) a xx 2^(2)` or `a = 1 m//s^(2)` From `upsilon = u + at = 0 + 1 xx 2 =2m//s` Impulse = change in momentum ` = m upsilon` `= 0.4 xx 2` `= 0.8 N-s` . |
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| 31. |
A body of mass 5 kg is moving with velocity of `v=(2hati+6hatj)ms^(-1)` at t=0s. After time t=2s, velocity of body is `(10 hati+6 hatj)`, then change in momentum to body isA. `40 hati kgms^(-1)`B. `20hati kgms^(-1)`C. `30 hati kgms^(-1)`D. `(50 hati+30 hatj)kgms^(-1)` |
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Answer» Correct Answer - A (a) Mass m= 5kg Change in velocity `v_(f)-v_(i)=[(10-2) hati+(6-6) hatj]` Change in momentum `=m Delta v=5[8hati]=40 hati kgms^(-1)` |
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| 32. |
If impulse/ varies with time t as f `(kg ms^(-1))=20t^(2)-40t`.The change in momentum is minimum atA. t=2sB. t=1sC. `t=(1)/(2)s`D. `t=(3)/(2)s` |
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Answer» Correct Answer - B (b) Impulse is defined as rate of change of momentum. For change in momentum to be minimum `(d)/(dt) ( 20t^(2)-400t)=0` 40t-40=0 t=1s |
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| 33. |
If in Q . 21 the speed of the stone is increased beyond the maximum permissible value and the string breaks suddenly which of the following correctly describes the trajectory of the stone after the string breaks : (a) the stone jerks radially outwards (b) the stone flies off tangentially from the instant the string breaks (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the stone ? |
| Answer» The instant the string breaks the stone flies off tangentially as per Newton s first law of motion . | |
| 34. |
A stone of of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of `40 rev//min` in a horizontal plane What is the tension in the string ? What is the maximum speed with which the stone can be whirled around If the string can withstand a maximum tension of 200 N ? |
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Answer» Here, m = 0.25 kg , r = 1.5 m n = 40 rpm = `40/60 rps = 2/3` rps Now T = mr`omega^(2)= mr(2pin)^(2)= 4pi^(2) mrn^(2)` T = `4 xx 22/7 xx 22/7 xx 0.25 xx 1.5 xx (2/2)^(2)` = 6.6 N If `T_max` = 200 N , then from , `T_max` = m`v_max^(2)/ r impliesv_max^(2)` = `T_max xx r / m` or `v_max^(2)= 200 x 1.5 / 0.25 = 1200impliessqrt 1200` = 34.6` ms^(-1)`. |
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| 35. |
A stone of of mass 0 .25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of `40 rev//min` in a horizontal plane What is the tension in the string ? What is the maximum speed with which the stone can be whirled around If the string can withstand a maximum tension of 200 N ? |
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Answer» Here `m = 0.25 kg , r = 1.5 m,n = 40 rpm = (40)/(60) "rps" = (2)/(3) rps , T = ?` `T = m t omega^(2) = m r (2 pi n )^(2) = 4 pi^(2) m r n^(2)` `T = 4 xx (22)/(7) xx (22)/(7) xx 0.25 xx 1.5 xx ((2)/(3))^(2) = 6.6 N` If `T_(max) = 200 N` then from `T_(max) = (m upsilon^(2)max)/(r)` `upsilon^(2)-_(max) = (T_(max) xx r)/(m) = (200 xx 1.5)/(0.25) = 1200` `upsilon_(max) = sqrt1200 = 34 .6 m//s`. |
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| 36. |
A motor cyclist loops a vertical loop of diameter 50 m , without dropping down even at uppermost point . What is the minimum speed at lowest and highest points of the loop ? |
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Answer» Here ` 50 // 2 = 25 m ` ` upsilon L = sqrt(5 r g )= sqrt( 5 xx 25 xx 9.8 ) = 35 m // s ` ` upsilon H = sqrt (rg )= sqrt (25 xx 9.8 ) = 15 . 65 m // s ` . |
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| 37. |
Why does a pilot not fall down when his aeroplane loops a vertical loop ? |
| Answer» This is because at the highest point of the (vertical) loop weight of the pilot is spent in providing the necessary centripetal force . | |
| 38. |
What is dry friction ? And what is wet friction ? |
| Answer» Dry friction is the force of friction between two solid surfaces in contact Wet friction is the force of friction between a solid surface and a liquid surface . | |
| 39. |
What is the unit of coefficient of friction depend ? |
| Answer» Coefficient of friction has no units . | |
| 40. |
An explosion blows a rock into three parts. Two parts go off at right angles to each other . These two are `1 kg` first part moving with a velocity of `12 ms^(-1) and 2 kg` second part moving with a velocity of `8 ms^(-1)`. If the third part flies off with a velocity of `4 ms^(-1)`. Its mass would beA. `3kg`B. `5kg`C. `7kg`D. `12kg` |
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Answer» Correct Answer - b Here, From `vec(p_(1)) + vec(p_(2)) + vec(p_(3)) = 0` `vec(p_(3)) = - (vec(p_(1)) +vec(p_(2)))` `|vec(p_(3))| = |vec(p_(1)) + vec(p_(2)) | =sqrt(p_(1)^(2) + p_(2)^(2))` `sqrt((1xx12)^(2) + (2xx8)^(2)) =20` `m_(3) = (|vecp_(3)|)/(upsilon_(3)) = (20)/(4) =5kg` . |
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| 41. |
An explosion blows a rock into three pieces Two pieces whose masses are 200 kg and 100 kg go off at `90^(@)` to eachother with a velocity of `8 m//s ` and ` 12 m//s ` respectively If the third piece flies off with a velocity of` 25 m//s` then calculate the mass of this piece and indicate the direction of flight of this piece in a diagram. |
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Answer» Here , ` m_(1) = 200 kg , upsilon_(1) = 8 m//s ` `m_(2) = 100 kg , upsilon_(2) = 12 m//s` `upsilon_(3) 25 m//s , m_(3) = ? ` `vec(p_(1)) = m_(1) vec(upsilon_(1)) = 200 xx 8 = 1600 kg m//s` , along OB `vec(p_(2)) = m_(2) vec (upsilon_(2)) = 100 xx 12 = 1200 kg m//s` , along OA Resultant momentum of these two pieces `p = sqrt(p_(1)^(2)+ p_(2)^(2)) = sqrt((1600)^(2) + (1200)^(2)` ` = 2000 kg m//s ` If `theta` is angle made by `vec(p)` with `vec(OA)` , then `tan theta = p_(1)/(p_(2)) = (1600)/(1200) = 1.33 ` `theta = tan^(-1) (1.33) = 53^(@)` From `p = m_(3) upsilon_(3)` `m_(3) = (p)/(upsilon_(3)) = (2000)/(25) = 80 kg ` |
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| 42. |
Two bodies A and B of mass 5 kg and 10 kg respectively in contact with eachother rest on a table against a rigid partition The coefficient of friction between the bodies and the table is 0.15 A force of 200 N is applied horizontally at A Calculate (i) reaction of the partition (ii) action reaction forces between A and B What happens when the partition is removed ? Does the anwer to (ii) change when the bodies are in motion ? Ignore difference between `mu_(s)` and `mu_(k)` . |
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Answer» Here , mass of body A, `m_(1) = 5 kg` mass of body B, `m_(2) = 10 kg` coefficient of friction between the bodies and the table ` mu = 0.15 ` Horizontal force applied on A , `F = 200 N ` (i) Force of limiting friction acting to the left `f = mu (m_(1) + m_(2) g` `= 0. 15 (5 + 10 ) xx 9.8 = 22 .05 N` Net force exerted on the partition (to the right) `F = F - f = 200 - 22.05 = 177.95 N ` Reaction of partition `= 177 . 95 N` to the left Force of limiting friction acting on the body A `f_(1) = mu m_(1) g = 0 .15 xx 5 xx 9.8 = 7 . 35 N` Net force exerted by body A on body B `F' = F - f_(1) = 200 - 7 .35 = 192 . 65 N` This force is to the right Reaction of body B on body A = 192 . 65 N to the left Acceleration produced in the system `a = (F)/(m_(1) + m_(2)) = (177 . 95)/(5 + 10) = 11 .86 m//s^(2)` Force producing motion in body A `F_(1) = m_(1) a = 5 xx 11.86 = 59.3 N` Net force exerted by body A on B when partition is removed ` = F ' - F_(1) = 192.65 - 59.3 = 133.35 N` Hence reaction of body B on body A when partition is removed `= 133 . 35 N` Thus answers to (ii) do change . |
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| 43. |
A body of 2 kg is suspended on a spring balance hung vertically in a lift . If the lift is falling downward under acceleration due to gravity g then what will be the reading of the balance ? If going upward with the same acceleration then ? |
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Answer» Reading of the balance = apparent weight ` = m (g-a) = m (g - g ) = zero ` If lift is going upwards with same acc , ` W = m (g - a ) = 2 m g = 4 kg` . |
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| 44. |
A force produces an acceleration `16m^(-2)` in a mass `0.5 kg` and an acceleration `4ms^(-2)` in an unknown mass when appied separately If both the masses are tied together what will be the acceleration under same force ? . |
| Answer» Correct Answer - `3.2 ms^(-2)`. | |
| 45. |
A force produces an acceleration of `4ms^(-2)` in a body of mass `m_(1)` and the same force produces an acceleration of `6ms^(-2)` in another body oa mass `m_(2)`. If the same force is applied to `(m_(1)+m_(2))`, then the acceleration will beA. `1.6ma^(-2)`B. `2ms^(-2)`C. `2.4ms^(-2)`D. `3.2ms^(-2)` |
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Answer» Correct Answer - c |
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| 46. |
A block is dragged on smooth plane with the help of a rope which moves with velocity v. the horizontal velocity of the block is A. vB. `(v)/(sin theta)`C. `v sin theta`D. `(v)/(cos theta)` |
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Answer» Correct Answer - d |
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| 47. |
A block slides down an incline of angle `30^(@)` with an acceleration of `g//4` Find the coefficient of kinetic friction . |
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Answer» Correct Answer - `1//2sqrt3` |
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| 48. |
A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will beA. `5"ms"^(-2)`B. `2"ms"^(-2)`C. `3.2"ms"^(-2)`D. `4.3"ms"^(-2)` |
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Answer» Correct Answer - C `"Here", r=10m, v=5ms^(-1), a_(t)=2m^(-2)` `a_(r)=(v^(2))/(r)=(5xx5)/(10)=2.5ms^(-2)` The net acceleration is `a=sqrt(a_(r)^(2)+a_(t)^(2))+sqrt((2.5)^(2)+2^(2))=sqrt(10.25)=3.2ms^(-2)` |
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| 49. |
A person of mass `60kg` is inside a lift of mass `940kg` and presses the button on control panel. The lift starts moving upword with an acceleration `1.0m//s^(2)` . If `g=10m//s^(2)` , the tension in the supporting cable is.A. 9680 NB. 11000 NC. 1200 ND. 8600 N |
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Answer» Correct Answer - B |
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| 50. |
A lift is moving upwards as shown in figure in figure A block of mass m s placed on a wedge inside the left and it is statioary with to lift Net force acting on m is A. zeroB. ma (upwards)C. `mg sintheta`D. ma (downwards) |
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Answer» Correct Answer - b |
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