39 + Interview Questions in FORCE-AND-NEWTONS-LAWS-OF-MOTION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	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.
Answer» As net force on the rod `=(F_(1)-F_(2))` and its mass is M. So, acceleration of the rod will be `a=(F_(1)-F_(2))//M` …(i) Now considering the motion of part AB of the rod [which has mass `(M//L)y` and acceleration a given by Eqn. (i)] assuming that tension at B is T, `F_(1)-T=(M//L)yxxa [from F=ma]` Substituting a from Eqn (i) `F_(1)-T=(M/Ly)((F_(1)-F_(2))/M)` or `T=F_(1)(1-y/L)+F_(2)(y/L)` To calculate tension at B we can also consider the motion of the other part of rod, i.e., BC. However, then equation of motion will be, `T-F_(2)=(M//L)(L-y)xxa` `"i.e " T=F_(2)+(M(L-y))/Lxx((F_(1)-F_(2)))/M` which of solving gives again, `T=F_(1)(1-y/L)+F_(2)(y/L)`

Discussion

2.	Two blocks each of mass M are resting on a frictionless inclined plane as shown in fig then: A. the block A moves down the planeB. the block B moves down the planeC. both the blocks remain at restD. both the blocks move down the plane
Answer» Correct Answer - A The force acting down the plane for blocks A and B are: `F_(A)=Mg sin 60^(@)=sqrt(3)/2 Mg=0.866Mg` `F_(B)=Mgsin 30^(@)=1/2Mg=0.5 Mg` Since `F_(A) gt F_(B)` the block A moves down the plane.

Discussion

3.	A 45 kg box starts from rest and moves on a smooth plane the force that varies with time as shown. The velocity of the box at t=8 sec is: A. `6 m//s`B. `8 m//s`C. `12 m//s`D. `24m//s`
Answer» Correct Answer - D `intF dt=M(v-u)="area"` `M(v-u)=1/2xx4^(2)xx180=360` `v=360/45=8m//sec` Velocity at 8 second `v_(f)=v+at=8+4xx4=24m//sec`

Discussion

4.	A particle is moving in a circular path. The acceleration and moment of the particle at a certain moment are `a=(4 hat(i)+3hat(j))m//s^(2) and p=(8 hat(i)-6hat(j))"kg-m/s"`. The motion of the particle isA. uniform circular motionB. accelerated circular motionC. decelerated circular motionD. we cannot say anything with `vec(a) and vec(p)` only
Answer» Correct Answer - B `vec(a)=4hati+3hatj` `vec(P)=8hat(i)-6hat(j)` `a_(c)=V^(2)/R=4` `a_(t)=((dV)/(dt))=3`

Discussion

5.	If a particle is compelled to mvoe on a given smooth plane curve under the action of given forces in the plane `vec(F)=xvec(i)+yvec(j),` then:A. `vec(F).dvec(r)=xdx+ydy`B. `intvec(F).dvec(r)ne1/2mv^(2)`C. `vec(F).dvec(r) nexdx+ydy`D. `1/2mv^(2)neint(xdx+ydy)`
Answer» Correct Answer - A `vec(F).dvec(r)=(xhat(i)+yhat(j)).(hat(i)dx+hat(j)dy)` `=xdx+ydy`

Discussion

6.	A particle moves in the x-y plane under is influence of a force such that its linear momentum is `vec(P)(t)=A[hat(i)cos (kt)-hat(j)sin (kt)]`, where A and k are constants Angle between the force and the momentum is:A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)`
Answer» Correct Answer - D `vec(F)=(dvec(P))/(dt)`

Discussion

7.	Fig shows a block of mass `0.1 kg` placed on a smooth wedge of mass `1/(5sqrt(3))` kg if the block of mass m will move vertically downward with acceleration `10m//s^(2)` Then the value of tension (in newton) in the string is `(theta=30^(@))` :
Answer» Correct Answer - 2 Accelerating to the given problem, m is a freely falling body, i.e., contact force between M and m is zero. Under this condition, acceleration of M leftwards will be `a=g cot alpha = g xx sqrt(3)` Tension `T=Ma=1/5sqrt(3)xxsqrt(3)g=2N`

Discussion

8.	A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.
Answer» Correct Answer - `((m_(1)-m_(0)+kt)g+kv_(0))/((m_(1)+m_(0)-kt))` Writing equations: `m_(1)g-T=m_(1)a` ...(i) `T-mg=ma` ....(ii) Adding equations (i) and (ii) `g(m_(1)-m)=(m_(1)+m)a` ....(iii) As `-(dm)/(dt)=k` `-underset(m_(0))overset(m)intdm=underset(0)overset(t)intkdt` `m_(0)-m=kt+C` `m_(0)-kt=m` By putting value of m in equation (iii) we get acceleration

Discussion

9.	If a body of mass m suspended by a spring comes to rest after a downward displacement `y_(0)` find (a) the force constant of the spring, (b) loss in gravitational potential energy and (c) gain in elastic potential energy.
Answer» (a) For equilibrium `ky_(0)=mg` i.e, `k=mg//y_(0)` (b) As the mass m has descended a distance `y_(0)` loss in potential energy `=mgy_(0)` (c) As the spring is stretched by `y_(0)` gain in elastic energy `=1/2ky_(0)^(2)=1/2mgy_(0)` [as from Eqn. (a)`k=(mg)/y_(0)`]

Discussion

10.	Block A and C start from rest and move to the right with acceleration `a_(A)=12t m//s^(2)` and `a_(C)=3m//s^(2)` Here t is in seconds. The time when block B attain comes to rest is: A. 2 sB. 1 sC. `3//2s`D. `1//2s`
Answer» Correct Answer - D `3=6t` `t=1/2`

Discussion

11.	In if the inclined plane is not accelerating and the block is then sliding down the plane, force exerted by the block on the plane is:A. `mg tan theta`B. `mg sin theta`C. mgD. none of these
Answer» Correct Answer - D `N=Mg cos theta`

Discussion

12.	Sand is being dropped on a conveyor belt at the rate of `Mkg//s` . The force necessary to kept the belt moving with a constant with a constant velocity of `vm//s` will be.A. `(Mv)/2` newtonB. zeroC. Mv newtonD. 2Mv newton
Answer» Correct Answer - C `F=vxx(dm)/(dt)=Mv`

Discussion

13.	Two bodies of masses 5 kg and 4 kg are arranged in two positions as shown in fig (a) and (b) if the pulleys and the table are perfectly smooth, the accelerations of the 5 kg body in case (a) and (b) are: A. `g and (5//9)g`B. `(4//9)g and (1//9)g`C. `g//5 and g//5`D. `(5//9)g and (1//9)g`
Answer» Correct Answer - B Case (a): `a_(1)=(4g)/((4+5))=4/9g` Case (b): `a_(2)=(5g-4g)/(5+4)=1/9g`

Discussion

14.	Two blocks of masses `m_(1)` and `m_(2)(m_(1) gt m_(2))` are connected by a massless threads, that passes over a massless smooth pulley. The pulley is suspended from the ceiling of an elevator. Now the elevator moves up with uniform velocity `v_(0)` Now, select the correct options: A. Magnitude of acceleration of `m_(1)` with respect to ground is greater than `((m_(1)-m_(2))g)/(m_(1)+m_(2))`B. Magnitude of acceleration of `m_(1)` with respect to ground is equal to `((m_(1)-m_(2))g)/(m_(1)+m_(2))`C. Tension in the thread that connects `m_(1)` and `m_(2)` is equal to `(2m_(1)m_(2)g)/(m_(1)+m_(2))`D. Tension in the thread that connects `m_(1)` and `m_(2)` is greater than `(2m_(1)m_(2)g)/(m_(1)+m_(2))`
Answer» Correct Answer - B::C The acceleration of the lift is zero `:.a=((m_(1)-m_(2))g)/((m_(1)+m_(2)))` `T=(2m_(1)m_(2)g)/((m_(1)+m_(2)))`

Discussion

15.	Two small balls of the same size and of masses `m_(1)` and `m_(2) (m_(1) gt m_(2))` are tied by a thin weightless thread and dropped from a balloon. The tension T of the thread during the flight towards the ground after the motion of the balls has become steady state is (steady state means that the balls are coming down with uniform velocity due to the forces weight, air friction and upthrust). Consider that air is still without any wind below, during the motion of the balls through air:A. zeroB. `((m_(1)-m_(2))g)/2`C. `((m_(1)+m_(2))g)/2`D. `m_(1)g`
Answer» Correct Answer - B `2T=(m_(1)-m_(2))g`

Discussion

16.	A particle tied to a string describes a vertical circular motion of radius r continually if it has a velocity `sqrt(3gr)` at the highest point, then the ratio of the respective tensions in the string holding it at the highest point and lowest points is:A. `4:3`B. `5:4`C. `1:4`D. `3:2`
Answer» Correct Answer - C Tension in the string at the highest point is, `T_(H)=(mv^(2))/r-mg=m/r(3gr)-mg=2mg` Tension in the string at eh lowest point is, `T_(L)=T_(H)+6mg=2mg+6mg=8mg` `:. T_(H)/T_(L)=(2mg)/(8mg)=1/4`

Discussion

17.	A solid body moves through air, at very high speed V faster than the velocity of molecules Show that the drag force on the body is proportional to `AV^(2)` where A is the frontal area of the body.
Answer» `F=Mxx(dV)/(dt)` `F=rhoV((dV)/dt)` `F=rhoA (dx)/(dt)xx(dV)=rhoAV^(2)` `F prop AV^(2)`

Discussion

18.	In a legend the hero kicked a baby pig so that he is projected with a speed greater than that of his cry. If the weight of the baby pig is assumed to be 5 kg and the time of contact `0.01` sec, the force with which the hero kicked him was:A. `5xx10^(-2)N`B. `2xx10^(5)N`C. `1.65xx10^(5)N`D. `1.65xx10^(3)N`
Answer» Correct Answer - B

Discussion

19.	Two billiard balls A and B each of mass 50 g and moving in opposite directions with speed of `5 m//s` each, collide and rebound with the same speed if the collision lasts for `10^(-3)` s which of the following statement(s) is (are) true?A. The impulse imparted to each ball is 0.25 kg `m//s` and te force on each ball is 250NB. The impulse imparted to each ball is 0.25 kg `m//s` and the force exerted on each ball is `25xx10^(-5)N`C. The impulse imparted to each ball is 0.5 ND. The impulse and the force on each ball are equal in magnitude and opposite in direction
Answer» Correct Answer - C::D `Ft=m(v-u)` `Fxx10^(-3)=50/1000[(5-(-5)]` `F=500N` Impuse and force on the two bells are equal and in the opposite directions.

Discussion

20.	The motion of a particle of mass m is given by x=0 for `t lt 0` s, x (t) = `A sin 4pit for 0 lt t(1//4) s (A gt 0)` and x=0 for `t gt (1//4)` s Which of the following statement(s) is (are) true?A. The force at `t=(1//8)` s on particle is `-16pi^(2)` A-mB. The particle is acted upon by an impulse of magnitude `(4pi^(2) A-m)` at t=0 s and `t=(1//4)` s.C. The particle is not acted upon by any forceD. The particle is not acted upon by a constant force
Answer» Correct Answer - A::B::D For `0 lt t lt 1/4` `x=Asin 4pit` `v=(dx)/(dt)=4piA cos4pit` `a=(dv)/(dt)=16pi^(2)A sin 4pit` At `t=1/8 s` `a=-16pi^(2)A sin4pixx1/8=-16pi^(2)A` `F=ma=-16pi^(2)A-m` Impulse = change in linear momentum `=Fxxt=-16pi^(2)A-mxx1/4=-4pi^(2)A-m` Impulse or change in linear momentum is the same at t=0 or `1/4` s Force F depends upon A which is not constant therefore particle is not acted upon by a constant force.

Discussion

21.	A body will not be in equilibrium if:A. `Sigmavec(F)=0 and Sigmavec(tau)=0`B. `Sigmavec(F)ne0 " but "Sigmavec(tau)=0`C. `Sigmavec(F)=0 " but "Sigmavec(tau)ne0`D. `Sigmavec(F)ne0 and Sigmavec(tau)ne0`
Answer» Correct Answer - B::C::D For translational equilibrium `Sigmavec(F)=0` For rotational equilibrium `Sigmavec(tau)=0`

Discussion

22.	A mass M is hung with a light inextensible string as shown in Find the tension in the horizontal part of the string .
Answer» Correct Answer - `sqrt(3)Mg` `T_(2) cos 60^(@)=Mg` `T_(2)=2Mg` ...(i) and for horizontal equilibium of P `T_(1)=T_(2) sin 60^(@)=T_(2)(sqrt(3)//2)` ...(ii) Substituting the value of `T_(2)` from eqn (i) in (ii) `T_(1)=(2Mg)xx(sqrt(3)//2)` `=(sqrt(3))Mg`

Discussion

23.	A strectching force of 1000 newton is applied at one end of a spring balance and an equal stretching force is applied at the other end at the same time. The reading of the balance will be:A. 2000 NB. zeroC. 1000 ND. 500 N
Answer» Correct Answer - C Net tension is zero

Discussion

24.	A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of:A. `(2//3)k`B. `(3//2)k`C. `3k`D. `6k`
Answer» Correct Answer - B for a spring `k prop 1/l` `1/k_(1)+1/k_(2)=1/k,` `:. 1/(2k_(1))+1/k_(2)=1/k {:[(l_(2)=2l_(1)),(":."k_(2)=k_(1)/2)]:}` `rArr k_(2)=3/2k`

Discussion

25.	A painter is applying force himself to raise him and the box with an acceleration of `5m//s^(2)` by a massless rope and pulley arrangement as shown in fig Mass of painter is 100 kg and that of box is 50 kg if `g=10 m//s^(2)` then: A. tension in the rope is 1125 NB. tension in the rope is 2250 NC. force of contact between the painter and the floor is 375 ND. none of these
Answer» Correct Answer - A::C For the system, the eqn. of motion is `2T-(M+m)g=(M+m)a` `2T=(M+m)(g+a)` `2T=(100+50)(10+5)` `T=1125 N` For the painter the eqn. of motion is `R+T=m(g+a)` `R+1125=100(10+5)` `R=375 N`

Discussion

26.	The velocity of a body of mass 20 kg decreases from `20m//s` to `5m//s` in a distance of 100 m. Force on the body is:A. `-27.5N`B. `-47.5N`C. `-37.5N`D. `-67.5N`
Answer» Correct Answer - C `F=M((v-u))/t` `(v^(2)=u^(2)-2as)`

Discussion

27.	In fig mass m is lifted up by attaching a mass 2m to the other end of the string while in fig (b) m is lifted by pulling the other end of the string with a constant force `F=2mg` then: A. acceleration of m in the two situations is same and non - zeroB. acceleration of m in situation (a) is more than that of in situation(b)C. acceleration of m in situation (a) is less than that of in situation (b)D. acceleration of m in the two situations is same and equals to zero
Answer» Correct Answer - C In situation (a) `2mg-T=2ma` `T-mg=ma` In situation (b) `T-mg=ma`

Discussion

28.	A body of mass `2kg` moving on a horizontal surface with an initial velocity of `4 ms^(-1)` comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. 8 NB. 4 NC. zeroD. 2 N
Answer» Correct Answer - B Retarding force `=-ma=-m(v-u)//t` `=2(4-0)//2=4N` :. To keep velcity uniform force required = opposing force =4N

Discussion

29.	Three blocks are connected as shown in the fig on a horizontal frictionless table if `m_(1)=1kg, m_(2)=8kg, m_(3)=27 kg` and `T_(3)=36N, T_(2)` will be: A. 18 NB. 9 NC. `3.375 N`D. `1.75 N`
Answer» Correct Answer - B `a=F//(m_(1)+m_(2)+m_(3))=36//(1+8+27)=1m//s^(2)` `T_(2)=(m_(1)+m_(2))a=(1+8)xx1=9N`

Discussion

30.	Three equal weights of mass 2 kg each are hanging on a string passing over a fixed pulley as shown in fig What is the tension in the string connecting weights B and C? A. zeroB. 13 NC. `3.3 N`D. `19.6 N`
Answer» Correct Answer - B `a=((m_(1)-m_(2))/(m_(1)+m_(2)))g=((2m-m)/(2m+m))g=g/3` As mass C is coming downward, `:. T=m(g-a)=2(g-g/3)=4/3xx10~=13N`

Discussion

31.	Find the tension `T_(2)` in the system shown in fig A. 1g NB. 2g NC. 5g ND. 6g N
Answer» Correct Answer - C `T_(2)=(M_(2)+M_(3))g=(2+3)gN=5gN`

Discussion

32.	In fig a sphere of mass 2 kg is suspended from the ceiling of a car which is initially at rest. Tension in the string in this situation is `T_(1)` The car now moves to the right with a uniform acceleration and the tension in the string is now `T_(2)` then: (Take `g=10m//s^(2)`) A. `T_(1)=T_(2)=20N`B. `T_(1)=20N, T_(2) gt20N`C. `T_(1)=20N, T_(2) lt20N`D. `T_(1) lt20N,T_(2)=20N`
Answer» Correct Answer - B `T_(1)=mg=2xx10=20N` `T_(2)=sqrt((mg)^(2)+(ma)^(2))=msqrt(g^(2)+a^(2))` Thus, `T_(2) gt T_(1)`

Discussion

33.	In a cricket match the fielder draws his hands backward after receiving the ball in order to take a catch because:A. his hands will be saved from getting hurtB. he deceives the playerC. it is a fashionD. he catches the ball firmly
Answer» Correct Answer - A To decrease force, time is increased

Discussion

34.	The length of an elastic string is x when tension is 5N. Its length is y when tension is 7N What will be its length when tension is 9N?A. `2y+x`B. `2y-x`C. `7x-4y`D. `7x+5y`
Answer» Correct Answer - B `kl=9` `l=(9/k)` `kx=5N` `ky=7N` `y=7/k`

Discussion

35.	A monkey of mass 30 kg climbs a rope which can withstand a maximum tension of 360 N. The maximum acceleration which this rope can tolerate for the climbing of monkey is: `(g=10m//s^(2))`A. `2 m//s^(2)`B. `3 m//s^(2)`C. `4 m//s^(2)`D. `5 m//s^(2)`
Answer» Correct Answer - A `T-Mg=Ma`

Discussion

36.	Calculate the volume of the balloon filled with hydrogen gas, which will be sufficient to lift a load of 25 kg in air. Given that the densities of air and hydrogen are `0.00129g//c c` and `0.00009 g//c c` respectively.
Answer» The weight of hydrogen in the balloon and load 25 kg will act vertically downwards while thrust of air vertically up. So the balloon will lift the weight if, `"Th" ge "weight of hydrogen"+"load"` or `Vrho_(air)g=Vrho_(H)g+Mg` or `V(rho_(air)-rho_(H))=M` or `(25xx10^(3)g)/((0.00129-0.00009))` `=(25xx10^(7))/12` or `V=2.083xx10^(7)"cc"` `=20.83m^(3)`

Discussion

37.	A load W is to be raised by a rope from rest to rest through a height h The greatest tension the rope can safely bear is nW. The least time in which the ascent can be made will be:A. `[(2h)/((n-1)g)]^(1//2)`B. `sqrt((2h)/g)`C. `[(2nh)/((n-1)g)]^(1//2)`D. `sqrt((2(n-1)h)/(ng))`
Answer» Correct Answer - C `h=1/2 at^(2)`

Discussion

38.	An elevator starts from rest with a constant upward acceleration it moves 2m in the first `0.6` second. A passenger in the elevator is holding a 3 kg package by a vertical string When the elevator is moving What is the tension in the string?A. 4 NB. `62.7 N`C. `29.4 N`D. `20.6 N`
Answer» Correct Answer - B `h=1/2at^(2)` `a=100/9 m//sec^(2)` `T=M(g+a)`

Discussion

39.	A person wishes to slide down a rope whose breaking load is `3/5` of the weight of the person Minimum acceleration by which the person should slide down without breaking the rope is:A. `0.8 g`B. `1.2 g`C. `0.6 g`D. `0.4 g`
Answer» Correct Answer - D `T=m(g-a)` `3/5 mg=m(g-a)` `3/5 g=g-a` `a=2/5g=0.4 g`

Discussion

Explore topic-wise InterviewSolutions in .

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.

Two blocks each of mass M are resting on a frictionless inclined plane as shown in fig then: A. the block A moves down the planeB. the block B moves down the planeC. both the blocks remain at restD. both the blocks move down the plane

A 45 kg box starts from rest and moves on a smooth plane the force that varies with time as shown. The velocity of the box at t=8 sec is: A. `6 m//s`B. `8 m//s`C. `12 m//s`D. `24m//s`

If a particle is compelled to mvoe on a given smooth plane curve under the action of given forces in the plane `vec(F)=xvec(i)+yvec(j),` then:A. `vec(F).dvec(r)=xdx+ydy`B. `intvec(F).dvec(r)ne1/2mv^(2)`C. `vec(F).dvec(r) nexdx+ydy`D. `1/2mv^(2)neint(xdx+ydy)`

A particle moves in the x-y plane under is influence of a force such that its linear momentum is `vec(P)(t)=A[hat(i)cos (kt)-hat(j)sin (kt)]`, where A and k are constants Angle between the force and the momentum is:A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)`

Fig shows a block of mass `0.1 kg` placed on a smooth wedge of mass `1/(5sqrt(3))` kg if the block of mass m will move vertically downward with acceleration `10m//s^(2)` Then the value of tension (in newton) in the string is `(theta=30^(@))` :

A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.

If a body of mass m suspended by a spring comes to rest after a downward displacement `y_(0)` find (a) the force constant of the spring, (b) loss in gravitational potential energy and (c) gain in elastic potential energy.

Block A and C start from rest and move to the right with acceleration `a_(A)=12t m//s^(2)` and `a_(C)=3m//s^(2)` Here t is in seconds. The time when block B attain comes to rest is: A. 2 sB. 1 sC. `3//2s`D. `1//2s`

In if the inclined plane is not accelerating and the block is then sliding down the plane, force exerted by the block on the plane is:A. `mg tan theta`B. `mg sin theta`C. mgD. none of these

Sand is being dropped on a conveyor belt at the rate of `Mkg//s` . The force necessary to kept the belt moving with a constant with a constant velocity of `vm//s` will be.A. `(Mv)/2` newtonB. zeroC. Mv newtonD. 2Mv newton

Two bodies of masses 5 kg and 4 kg are arranged in two positions as shown in fig (a) and (b) if the pulleys and the table are perfectly smooth, the accelerations of the 5 kg body in case (a) and (b) are: A. `g and (5//9)g`B. `(4//9)g and (1//9)g`C. `g//5 and g//5`D. `(5//9)g and (1//9)g`

A particle tied to a string describes a vertical circular motion of radius r continually if it has a velocity `sqrt(3gr)` at the highest point, then the ratio of the respective tensions in the string holding it at the highest point and lowest points is:A. `4:3`B. `5:4`C. `1:4`D. `3:2`

A solid body moves through air, at very high speed V faster than the velocity of molecules Show that the drag force on the body is proportional to `AV^(2)` where A is the frontal area of the body.

A body will not be in equilibrium if:A. `Sigmavec(F)=0 and Sigmavec(tau)=0`B. `Sigmavec(F)ne0 " but "Sigmavec(tau)=0`C. `Sigmavec(F)=0 " but "Sigmavec(tau)ne0`D. `Sigmavec(F)ne0 and Sigmavec(tau)ne0`

A mass M is hung with a light inextensible string as shown in Find the tension in the horizontal part of the string .

A strectching force of 1000 newton is applied at one end of a spring balance and an equal stretching force is applied at the other end at the same time. The reading of the balance will be:A. 2000 NB. zeroC. 1000 ND. 500 N

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of:A. `(2//3)k`B. `(3//2)k`C. `3k`D. `6k`

The velocity of a body of mass 20 kg decreases from `20m//s` to `5m//s` in a distance of 100 m. Force on the body is:A. `-27.5N`B. `-47.5N`C. `-37.5N`D. `-67.5N`

A body of mass `2kg` moving on a horizontal surface with an initial velocity of `4 ms^(-1)` comes to rest after `2` second. If one wants to keep this body moving on the same surface with a velocity of `4 ms^(-1)` the force required isA. 8 NB. 4 NC. zeroD. 2 N

Three blocks are connected as shown in the fig on a horizontal frictionless table if `m_(1)=1kg, m_(2)=8kg, m_(3)=27 kg` and `T_(3)=36N, T_(2)` will be: A. 18 NB. 9 NC. `3.375 N`D. `1.75 N`

Three equal weights of mass 2 kg each are hanging on a string passing over a fixed pulley as shown in fig What is the tension in the string connecting weights B and C? A. zeroB. 13 NC. `3.3 N`D. `19.6 N`

Find the tension `T_(2)` in the system shown in fig A. 1g NB. 2g NC. 5g ND. 6g N

In a cricket match the fielder draws his hands backward after receiving the ball in order to take a catch because:A. his hands will be saved from getting hurtB. he deceives the playerC. it is a fashionD. he catches the ball firmly

The length of an elastic string is x when tension is 5N. Its length is y when tension is 7N What will be its length when tension is 9N?A. `2y+x`B. `2y-x`C. `7x-4y`D. `7x+5y`

A monkey of mass 30 kg climbs a rope which can withstand a maximum tension of 360 N. The maximum acceleration which this rope can tolerate for the climbing of monkey is: `(g=10m//s^(2))`A. `2 m//s^(2)`B. `3 m//s^(2)`C. `4 m//s^(2)`D. `5 m//s^(2)`

Calculate the volume of the balloon filled with hydrogen gas, which will be sufficient to lift a load of 25 kg in air. Given that the densities of air and hydrogen are `0.00129g//c c` and `0.00009 g//c c` respectively.

A load W is to be raised by a rope from rest to rest through a height h The greatest tension the rope can safely bear is nW. The least time in which the ascent can be made will be:A. `[(2h)/((n-1)g)]^(1//2)`B. `sqrt((2h)/g)`C. `[(2nh)/((n-1)g)]^(1//2)`D. `sqrt((2(n-1)h)/(ng))`

An elevator starts from rest with a constant upward acceleration it moves 2m in the first `0.6` second. A passenger in the elevator is holding a 3 kg package by a vertical string When the elevator is moving What is the tension in the string?A. 4 NB. `62.7 N`C. `29.4 N`D. `20.6 N`

A person wishes to slide down a rope whose breaking load is `3/5` of the weight of the person Minimum acceleration by which the person should slide down without breaking the rope is:A. `0.8 g`B. `1.2 g`C. `0.6 g`D. `0.4 g`