619 + Interview Questions in NEWTONS-LAWS-OF-MOTION IN GENERAL AWARENESS Page 4 InterviewSolution

151.	A vehicle of mass `m` is moving on a rough horizontal road with momentum `P` . If the coefficient of friction between the tyres and the road br mu, then the stopping distance is:A. `(P)/(2mumg)`B. `(P^(2))/(2mumg)`C. `(P)/(2mum^(2)g)`D. `(P^(2))/(2mum^(2)g)`
Answer» Correct Answer - D `S=(u^(2))/(2mug)=(m^(2)-u^(2))/(2mumg^(2))=(P^(2))/(2mumg^(2)g)`

Discussion

152.	A body of mass 100g is sliding on a inclined plane with an inclination of `60^(@)`. What is the frictional force experienced, if coeficient of friction is 1.7? (Take `g=10m//s^(2)`)A. 0.85NB. 0.95NC. 1.05ND. 1.145N
Answer» Correct Answer - A `f=1.7xx(100)/(1000)xx10xx(1)/(2)=0.85N`

Discussion

153.	Assertion: Friction is a self-adjusting force. Reason: A block on the horizontal table is acted upon by a force F. The graph of frictional force against applied force F is. A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false
Answer» Correct Answer - D Graph explain self adjusting value

Discussion

154.	Assertion: Static frictional force is a self adjusting force. Reason: Force of static friction does not depend upon the mass of the body.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false
Answer» Correct Answer - B `f_(s)("max")=muN` depends on `mu` and N and not on mass.

Discussion

155.	Assertion: When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of tendenc of motion his leg. Reason: If a body moves with constant velocity, then it means that body is moving along straight line `&` there is no acceleration.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false
Answer» Correct Answer - B Reason do not explain assertion.

Discussion

156.	Assertion: Frictional force is the component of contact force parallel to the surface. Reason: Friction force always opposes the motion of a body.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false
Answer» Correct Answer - C Friction force opposes relative motion.

Discussion

157.	A turn of radius `20 m` is banked for the vehicle of mass `200 kg` going at a speed of `10 m//s`. Find the direction and magnitude of frictional force (a) `5 m//s` (b) `15 m//s` Assume that friction is sufficient to prevent slipping. `(g=10 m//s^(2))`
Answer» `v =10m//s` `tan theta = (v^(2))/(rg) = (10)^(2)/((20)(10))=(1)/(2)rArrtheta=tan^(-1) ((1)/(2))` Now as speed is decreased force of friction acts upwards Using the equations `N sin theta -f cos theta = (mv^(2))/(R ) , N cos theta + f sin theta = mg` Substituting `theta = tan^(-1) ((1)/(2)), v =5m//s, m =-200kg` and `r =20m` in the above equations, we get `f =300 sqrt5N` ltbr gt(b) In the second case force of friction f will act downwards `N sin theta + f cos theta = (mv^(2))/(r) , N cos theta -f sin theta = mg` Substituting `theta = tan^(-1) ((1)/(2)), v =15m//s` `m= 200kg` and `r =20m`, in the above equations we get `f =500 sqrt5 N` .

Discussion

158.	Assertion: Magnitude of the contact force is always greater than the magnitude of frictional force. Reason: Contact force is the resultant of the friction force and normal reaction.A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.C. If Assertion is Trie but the Reason is False.D. If both Assertion `&` Reason are false
Answer» Correct Answer - A `F_(C)=sqrt(f^(2)+N^(2))`

Discussion

159.	A 10 kg mass is resting on a horizontal surface and horizontal force of 80N is applied. If `mu = 0.2`, the ratio of acceleration without and with frication is `(g = 10 ms)^(2)`A. `3//4`B. `4//3`C. `1//2`D. `2`
Answer» Correct Answer - B `F_("net") = ma, a_(1) = (F)/(m),a_(2) =(F-mu_(k)mg)/(m)`

Discussion

160.	A body of mass `5xx10^(-3)` kg is launched upon a rough inclined plane making an angle of `30^(@)` with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.
Answer» Correct Answer - [0.36]

Discussion

161.	Two blocks of mass 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length 1 meter, see fig. The upper wire has negligible mass and the lower wire has a uniform mass of `0.2kg//m`. The whole system of blocks wires and support have an upward acceleration of `0.2m//s^2`. Acceleration due to gravity is `9.8m//s^2`. (i) Find the tension at the mid-point of the lower wire. (ii) Find the tension at the mid-point of the upper wire.
Answer» Correct Answer - [20 N, 50 N]

Discussion

162.	To avoid slipping while walking on ice, one should take smaller steps because of theA. larger frictionB. smaller frictionC. larger normal forceD. smaller normal force
Answer» Correct Answer - C

Discussion

163.	Three equal weight `A,B` and `C` of mass `2kg` each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights `B` and `C` is approximately A. ZeroB. 13 NC. `3.3` ND. `19.6` N
Answer» Correct Answer - B

Discussion

164.	A satellite in a force - free space sweeps stationary interplanetary dust at a rate `dM//dt = alpha v`, where `M` is the mass , `v`is the velocity of the satellite and `alpha` is a constant. What is the deacceleration of the satellite ?A. `-2 alphav^(2)//M`B. `-2alphav^(2)//M`C. `alphav^(2)//M`D. `-alphav^(2)`
Answer» Correct Answer - C `F=-alphav^(2)` `therefore a=(-alphav^(2))/(M)`

Discussion

165.	A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block will beA. `p`B. `(Pm)/(M+m)`C. `(PM)/(M+m)`D. `(Pm)/(M-m)`
Answer» Correct Answer - C

Discussion

166.	A block of mass `M` is pulled along a horizontal frictionless surface by a rope of mass `m` . Force P is applied at one end of rope. The force which the rope exerts on the block is:A. F/(M+m)B. F should be equal to the weight of A and BC. FM/(m+M)D. Zero
Answer» Correct Answer - C `a=(F)/(M+m)` Force on `M=Mxx(F)/(M+m)`

Discussion

167.	A horizontal uniform rope of length L, resting on a frictionless horizontal surface, is pulled at one end by force F. What is the tension in the rope at a distance l from the end where the force is applied?A. `(F(y-x))/(y)`B. `(F.y)/(y-x)`C. `(F.y)/(x)`D. `(F.y)/(y)`
Answer» Correct Answer - A

Discussion

168.	Let `vecu` be the initial velocity of a particle and `vecF` be the resultant force acting on it . Describe the path that the particle can take if (a) `vecu xx vecF = 0 "and" vecF = "constant"` (b) `vecu.vecF = 0 "and" vecF = "constant"` In which case can the particle retrace its path .
Answer» Correct Answer - (a) straight line (b) Parabolic

Discussion

169.	In the arrangement shown in figure pulley P can move whereas other two pulleys are fixed. All of them are light. String is light and inextensible. The coefficient of friction between 2 kg and 3 kg block is `mu = 0.75` and that between 3 kg block and the table is `mu = 0.5`. The system is released from rest (i) Find maximum value of mass M, so that the system does not move. Find friction force between 2 kg and 3 kg blocks in this case. (ii) If M = 4 kg, find the tension in the string attached to 2 kg block. (iii) If M = 4 kg and `mu_(1) = 0.9`, find friction force between the two blocks, and acceleration of M. (iv) Find acceleration of M if `m_(1) = 0.75, m_(2) = -0.9` and M = 4 kg.
Answer» Correct Answer - (i) 2.5 kg ; 12.5 N (ii) `(50)/(3) N ` (iii) `(40)/(3) N , a = (5)/(3) m//s^(2)` (iv) `(5)/(6) m//s^(2)`

Discussion

170.	Two masses m and m `(mgtm)` are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass isA. `((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g`B. `(m_(1)-m_(2))/(m_(1)+m_(2))g`C. `(m_(1)+m_(2))/(m_(1)-m_(2))g`D. Zero
Answer» Correct Answer - B

Discussion

171.	A lift goes up with 10 m/s. a pulley P is fixed to the ceiling of the lift. To this pulley other two pulley `P_(1)` and `P_(2)` are attached. `P_(1)` moves up with velocity 30 m/s. A moves up with velocity 10 m/s. D is moving downwards with velocity 10 m/s at same instant of time. Find the velocity of B and that of C at that instant. Assume that all velocities are relative to the ground.
Answer» Correct Answer - 5 Apply constraint on pulley `P` `vecv_(p_(1)//p)=-vecv_(p_(2)//p)),vecv_(p_(1))-vecv_(p)=-(vecv_(p_(2))-vecv_(p))` `vecv_(p_(1)),vecv_(p),vecv_(p_(2))` are respective velocity w.r.t ground `vecv_(p_(2))=2vecv_(p)-vecv_(p_(1)),=2[10hatj]-[-30hatj]=-10hatj` Now apply constraint eqn on pulley `P_(2)` `vecvc-vecv_(p_(2))=-(vecvp-vp_(2))` `vecvc=2vecv_(p_(2))-vecv_(D)=2[-10hatj]-[-10hatj]vecv_(C)=10hatj` Apply constraint eqn on pulley `P_(1)` to get `vecv_(A)-vecv_(p_(1))=-(vecv_(B)-vecv_(p_(1))),vecv_(B)=2vecv_(p_(1))-vecv_(A)` `=2[30hatj]-[10hatj]=50hatj` .

Discussion

172.	In the arrangement shown, the pulleys are smooth and the strings are inextensible. The acceleration of block B is : A. g/5B. 5g/5C. 2g/5D. 2g/3
Answer» Correct Answer - C

Discussion

173.	The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle `theta` should be A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)`
Answer» Correct Answer - C

Discussion

174.	In the arrangement shown in the Fig, the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed A. `V cos theta`B. `V // cos theta`C. `2V cos theta`D. `2//V cos theta`
Answer» Correct Answer - B ` U cos theta=V` `therefore U=(v)/(cos theta)`

Discussion

175.	A lift can move upward or downward. A light inextensible string fixed from ceiling of lift when a frictionless pulley and tensions in string `T_(1)`. Two masses of `m_(1) and m_(2)` are connected with inextensible light string and tension in this string `T_(2)` as shown in figure. Read the questionbs carefully and answer If `m_(1)=m_(2) and m_(1)` is moving at a certain instant with velocity `v` upward with respect to lift and the lift is moving in upward direction with constant acceleration `(a lt g)` then speed of `m_(1)` with respect to lift :A. `T_(1):T_(2):T_(3)::1:1:1`B. `T_(1):T_(2):T_(3)::6:5:3`C. `T_(1):T_(2):T_(3)::3:5:5`D. `T_(1):T_(2):T_(3)::6:5:6`
Answer» Correct Answer - B `T_(1) mg=mxx2 thereforeT_(1)=mxx12` `T_(2)=mg T_(2)=mxx10` `mg-T_(3)=mxx4 T_(3)=mxx16` `T_(1):T_(2):T_(3)=6:5:3`

Discussion

176.	Two bodies of masses 5 kg and 4 kg are arranged in two possition as shown in fig. (A) and (B), if the pulleys and the table are perfectly smooth, the accelration of the 5 kg. body in case (A) and (B) are:- A. g and (5/9)gB. (4/9)g and (1/9)gC. g/5 and g/5D. (5/9)g and (1/9)g
Answer» Correct Answer - B For case A `4g-T=4a` T=5a `therefore a=(4g)/(9)` For case B 5g-T=5a T-4g=4a `a=(g)/(9)`

Discussion

177.	There blocks of masses `m_(1)` , `m_(2)` and `m_(3)` are connected by may less unstretchable strings on a smooth surface. Tension `T_(2)` is. A. `10N`B. `20N`C. `32N`D. `40N`
Answer» Correct Answer - C `F_("net") =ma, T_(3) -T_(2) =m_(3) a, T_(2) -T_(1) =m_(2) a,T_(1) =m_(1)a` solving the above equation, we get .

Discussion

178.	There blocks of masses `m_(1)` , `m_(2)` and `m_(3)` are connected by may less unstretchable strings on a smooth surface. Tension `T_(2)` is. A. `2.3N`B. `32N`C. `23N`D. `3.2N`
Answer» Correct Answer - B Aceleration, `a=(F)/(m_(1)+m_(2)+m_(3))` `=(40)/(10+6+4)=2ms^(-2)` Tension `T_(2)=(m_(1)+m_(2))a` `=(10=6)2=32N` .

Discussion

179.	Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B. Suppose their exerts a constant and equal force of resistance on the two bodies.A. The two bodies will reach the same heightB. A will go higher that BC. B will go higher than AD. Any of the above three may happen depending on the speed with which the object are thrown.
Answer» Correct Answer - B

Discussion

180.	Three blocks of masses `m_(1),m_(2)`and `m_(3)` are connected by massless strings as shown on a frictionless table. They are pulled with a force `T_(3)=40N`. If `m_(1)=10kg,m_(2)=6kg` and `m_(3)=4kg` the tension `T_(2)` will be A. 10NB. 20NC. 32ND. 40N
Answer» Correct Answer - C The common acceleration of the blocks is `a (40)/(10+6+4)=(40)/(20)=2 m//s^(2)` `therefore` Tension `T_(2)=(m_(1)+m_(2))a=(10+6)2` `=12xx2=32N`.

Discussion

181.	Three blocks of masses `m_(1),m_(2)`and `m_(3)` are connected by massless strings as shown on a frictionless table. They are pulled with a force `T_(3)=40N`. If `m_(1)=10kg,m_(2)=6kg` and `m_(3)=4kg` the tension `T_(2)` will be A. 20 NB. 40 NC. 10 ND. 32 N
Answer» Correct Answer - D

Discussion

182.	The system shown in figure is in equilibrium. Pulley, springs and the strings are massless. The three blocks A, B and C have equal masses. `x_(1)` and `x_(2)` are extensions in the spring 1 and spring 2 respectively. (a) Find the value of `\|(d^(2)x^(2))/(dt^(2))\|` immediately after spring 1 is cut . (b) Find the value of `\|(d^(2)x^(1))/(dt^(2))\|` and `\|(d^(2)x^(2))/(dt^(2))\|` immediately after string AB is cut. (c) Find the value of `\|(d^(2)x^(1))/(dt^(2))\|` and `\|(d^(2)x^(2))/(dt^(2))\|` immediately after spring 2 is cut .
Answer» Correct Answer - (a) `\|(d^(2)x_(2))/(dt^(2))\| = (3g)/(2)` `\|(d^(2)x_(1))/(dt^(2))\|= 2g ; \|(d^(2)x_(2))/(dt^(2))\| = 2g` (c) `\|(d^(2)x_(1))/(dt^(2))\| = (g)/(2) ; \|(d^(2)x_(2))/(dt^(2))\| = (3g)/(2)`

Discussion

183.	Two masses `m_(1)` and `m_(2)` are attached to a string which passes over a frictionless smooth pulley. When `m_(1)=10kg,m_(2)=6kg`, the acceleration of masses is A. `20m//s^(2)`B. `5m//s^(2)`C. `2.5m//s^(2)`D. `10m//s^(2)`
Answer» Correct Answer - C

Discussion

184.	Two masses `m_1=5kg` and `m_2=4.8kg` tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move? A. `0.2m//s^(2)`B. `9.8m//s^(2)`C. `5m//s^(2)`D. `4.8m//s^(2)`
Answer» Correct Answer - A

Discussion

185.	Two blocks of masses `m_(1)` and `m_(-2)mgt (m_(2))` are connected by 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 greate than `((m_(1)-m_(2))g)/(m_(1)+m_(2))`B. Magnitude of acceleration of `m_(1)` with respect to ground is greate than `((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 equal to `(2m_(1)m_(2)g)/(m_(1)+m_(2)` .
Answer» Correct Answer - B::C Draw `FBD` and use Newtons equations .

Discussion

186.	Ten one-rupee coins are put on top each other on a table. Each coin has a mass m. The rection of the `6^(th)` coin (counted from the bottom) on the `7^(th)` coin isA. 3 mgB. 7mgC. 2mgD. 5mg
Answer» Correct Answer - A 3mg

Discussion

187.	A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. If the ball was in contact with the floor for 0.01s. What was its average acceleration during contact? `(g=10m//s^(2))`A. 300 `m//s^(2)`B. `2000 m//s^(2)`C. `1000 m//s^(2)`D. `500m//s^(2)`
Answer» Correct Answer - A `F_("avg")=(msqrt(2+10+10)+msqrt(2xx10xx5))/(0.01)` `therefore a_("avg")=(10(sqrt(2)+1))/(0.01)=3000m//s^(2)`

Discussion

188.	A ball weighing `10g` hits a hard surface vertically with a speed of `5m//s` and rebounds with the same speed The ball remains in contact with the surface speed The ball remains in contact with the surface for `0.01s` The average force exerted by the surface on the ball is .A. 100 NB. 10 NC. 1 ND. 150 N
Answer» Correct Answer - B

Discussion

189.	In the previous problem, If the string C is stretched slowly, thenA. The portion AB of the string will breakB. The portion BC of the string will breakC. None of the string will breakD. None of the above
Answer» Correct Answer - A When the spring `C` is stretched slowly, the tension in `A` is greater than that of `C` , because of the weigth mg and the former reaches breacking point earlier.

Discussion

190.	Solve the previous problem if the blocks A and B are of different shapes as shown in figure-2.203.
Answer» Correct Answer - `[((f-3mg sin theta)/(M+10m))sqrt((10-6costheta))]`

Discussion

191.	In the previous problem, the height to which the lift takes the passenger isA. 50 mB. 60 mC. 70 mD. 78 m
Answer» Correct Answer - D Displacement = area of v-t graph `=(1)/(2)(3+10)xx12=78m`

Discussion

192.	In the previous problem if `M=80kg` and `m=410kg` and the boy pulls the rope such that the system (boy+box) moves upward with acceleration `5m//s^(2)`,find the force exerted by the boy and contact force between the boy and the box.
Answer» Boy `:` `N_(0)-Mg=Ma` `rArrT+N_(0)=M(g+a)=80(10+5)=1200` (i) Box `:` `T-N_(0)-mg=ma` `rArrT-N_(0)=m(g+a)=40(10+5)=600` (ii) `T=900N,N_(0)=300N`

Discussion

193.	A bullet of mass 40g moving with a speed of `90ms^(-1)` enters a heavy wooden block and is stopped after a direction of 60cm. The average resistive force exered by the block on the bullet isA. 180NB. 220NC. 270ND. 320N
Answer» Correct Answer - C Here, `u=90ms^(-1)` , `v=0` . `m=40g(40)/(1000)kg=0.04kg` . `s=60cm=0.6m` . Using `v^(2)-u^(2)=2as` . :. `(0)^(2)-(90)^(2)=2axx0.6` . `:. A=(90)^(2)/(2xx0.6)=-6750ms^(-2)` . `-ve` sign shows the retardation. The average resistive force exerted by block on the bullet is. `F=mxxa=(0.04kg)(6750ms^(-2))=270N` .

Discussion

194.	A `30gm` bullet initially at `120m//s` penetrates `12cm` into a wooden block. The average block. The average resistance exerted by the wooden block is.A. `2850N`B. `2200N`C. `2000N`D. `1800N`
Answer» Correct Answer - D `F=(m(u^(2)-v^(2)))/(2s)=(30xx10^(-3)xx(120)^(2))/(2xx12xx10^(-2))=1800N`

Discussion

195.	A monkey of mass `20kg` is holding a vertical rope. The rope will not break when a mass of `25kg` is suspended from it but will break it the mass exeeds `25kg` . What is the maximum acceleration with which the monkey can climb up along the rope? `(g=10m//s^(2))` .A. `10m//s^(2)`B. `25m//s^(2)`C. `2.5m//s^(2)`D. `5m//s^(2)`
Answer» Correct Answer - C Tension the string `=m(g+a)` =Breacking force implies `20(g+a)=25xxg` implies `a=g//4=2.5m//s^(2)`

Discussion

196.	In the previous problem, If the string C is stretched slowly, thenA. The portion AB of the string will breakB. The portion BC of the string will breakC. None of the strings will breakD. None of the above
Answer» Correct Answer - A

Discussion

197.	If a bullet of mass 5gm moving with velocity `100m/sec, penertates the wooden block upto 6cm. Then the average force imposed by the bullet on the block isA. 8300 NB. 417 NC. 830 ND. Zero
Answer» Correct Answer - B

Discussion

198.	When a bicycle is motion the force of friction exerted by the ground on the two wheels is such that is acts .A. in the backward direction on the front wheel and in the forward direaction on the rear wheel .B. in the backward direction on the front wheel and in the forward direaction on the rear wheel .C. in the backward direction on both the front and rear wheelsD. in the forward direction on the front and rear wheels
Answer» Correct Answer - A

Discussion

199.	When a bicycle is motion the force of friction exerted by the ground on the two wheels is such that is acts .A. In the backward direction on the front wheel and in the forward direction on the rear wheel.B. In the forward direction on the front wheel and in the backward direction on the rear wheel.C. In the backward direction on both the front and on the rearwheel.D. In the forward direction on both the front and on the rear wheel
Answer» Correct Answer - A::C

Discussion

200.	A block of metal weighing 2kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1kg/s and at a speed of 5 m/s. Calculate the initial acceleration of the block.
Answer» Correct Answer - `[2.5 m//s^(2)]`

Discussion

Explore topic-wise InterviewSolutions in .

A vehicle of mass `m` is moving on a rough horizontal road with momentum `P` . If the coefficient of friction between the tyres and the road br mu, then the stopping distance is:A. `(P)/(2mumg)`B. `(P^(2))/(2mumg)`C. `(P)/(2mum^(2)g)`D. `(P^(2))/(2mum^(2)g)`

A body of mass 100g is sliding on a inclined plane with an inclination of `60^(@)`. What is the frictional force experienced, if coeficient of friction is 1.7? (Take `g=10m//s^(2)`)A. 0.85NB. 0.95NC. 1.05ND. 1.145N

A turn of radius `20 m` is banked for the vehicle of mass `200 kg` going at a speed of `10 m//s`. Find the direction and magnitude of frictional force (a) `5 m//s` (b) `15 m//s` Assume that friction is sufficient to prevent slipping. `(g=10 m//s^(2))`

A 10 kg mass is resting on a horizontal surface and horizontal force of 80N is applied. If `mu = 0.2`, the ratio of acceleration without and with frication is `(g = 10 ms)^(2)`A. `3//4`B. `4//3`C. `1//2`D. `2`

A body of mass `5xx10^(-3)` kg is launched upon a rough inclined plane making an angle of `30^(@)` with the horizontal. Obtain the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

To avoid slipping while walking on ice, one should take smaller steps because of theA. larger frictionB. smaller frictionC. larger normal forceD. smaller normal force

Three equal weight `A,B` and `C` of mass `2kg` each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights `B` and `C` is approximately A. ZeroB. 13 NC. `3.3` ND. `19.6` N

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block will beA. `p`B. `(Pm)/(M+m)`C. `(PM)/(M+m)`D. `(Pm)/(M-m)`

A block of mass `M` is pulled along a horizontal frictionless surface by a rope of mass `m` . Force P is applied at one end of rope. The force which the rope exerts on the block is:A. F/(M+m)B. F should be equal to the weight of A and BC. FM/(m+M)D. Zero

A horizontal uniform rope of length L, resting on a frictionless horizontal surface, is pulled at one end by force F. What is the tension in the rope at a distance l from the end where the force is applied?A. `(F(y-x))/(y)`B. `(F.y)/(y-x)`C. `(F.y)/(x)`D. `(F.y)/(y)`

Let `vecu` be the initial velocity of a particle and `vecF` be the resultant force acting on it . Describe the path that the particle can take if (a) `vecu xx vecF = 0 "and" vecF = "constant"` (b) `vecu.vecF = 0 "and" vecF = "constant"` In which case can the particle retrace its path .

Two masses m and m `(mgtm)` are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass isA. `((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g`B. `(m_(1)-m_(2))/(m_(1)+m_(2))g`C. `(m_(1)+m_(2))/(m_(1)-m_(2))g`D. Zero

In the arrangement shown, the pulleys are smooth and the strings are inextensible. The acceleration of block B is : A. g/5B. 5g/5C. 2g/5D. 2g/3

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle `theta` should be A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)`

In the arrangement shown in the Fig, the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed A. `V cos theta`B. `V // cos theta`C. `2V cos theta`D. `2//V cos theta`

Two bodies of masses 5 kg and 4 kg are arranged in two possition as shown in fig. (A) and (B), if the pulleys and the table are perfectly smooth, the accelration of the 5 kg. body in case (A) and (B) are:- A. g and (5/9)gB. (4/9)g and (1/9)gC. g/5 and g/5D. (5/9)g and (1/9)g

There blocks of masses `m_(1)` , `m_(2)` and `m_(3)` are connected by may less unstretchable strings on a smooth surface. Tension `T_(2)` is. A. `10N`B. `20N`C. `32N`D. `40N`

There blocks of masses `m_(1)` , `m_(2)` and `m_(3)` are connected by may less unstretchable strings on a smooth surface. Tension `T_(2)` is. A. `2.3N`B. `32N`C. `23N`D. `3.2N`

Three blocks of masses `m_(1),m_(2)`and `m_(3)` are connected by massless strings as shown on a frictionless table. They are pulled with a force `T_(3)=40N`. If `m_(1)=10kg,m_(2)=6kg` and `m_(3)=4kg` the tension `T_(2)` will be A. 10NB. 20NC. 32ND. 40N

Three blocks of masses `m_(1),m_(2)`and `m_(3)` are connected by massless strings as shown on a frictionless table. They are pulled with a force `T_(3)=40N`. If `m_(1)=10kg,m_(2)=6kg` and `m_(3)=4kg` the tension `T_(2)` will be A. 20 NB. 40 NC. 10 ND. 32 N

Two masses `m_(1)` and `m_(2)` are attached to a string which passes over a frictionless smooth pulley. When `m_(1)=10kg,m_(2)=6kg`, the acceleration of masses is A. `20m//s^(2)`B. `5m//s^(2)`C. `2.5m//s^(2)`D. `10m//s^(2)`

Two masses `m_1=5kg` and `m_2=4.8kg` tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move? A. `0.2m//s^(2)`B. `9.8m//s^(2)`C. `5m//s^(2)`D. `4.8m//s^(2)`

Ten one-rupee coins are put on top each other on a table. Each coin has a mass m. The rection of the `6^(th)` coin (counted from the bottom) on the `7^(th)` coin isA. 3 mgB. 7mgC. 2mgD. 5mg

A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. If the ball was in contact with the floor for 0.01s. What was its average acceleration during contact? `(g=10m//s^(2))`A. 300 `m//s^(2)`B. `2000 m//s^(2)`C. `1000 m//s^(2)`D. `500m//s^(2)`

In the previous problem, If the string C is stretched slowly, thenA. The portion AB of the string will breakB. The portion BC of the string will breakC. None of the string will breakD. None of the above

Solve the previous problem if the blocks A and B are of different shapes as shown in figure-2.203.

In the previous problem, the height to which the lift takes the passenger isA. 50 mB. 60 mC. 70 mD. 78 m

In the previous problem if `M=80kg` and `m=410kg` and the boy pulls the rope such that the system (boy+box) moves upward with acceleration `5m//s^(2)`,find the force exerted by the boy and contact force between the boy and the box.

A bullet of mass 40g moving with a speed of `90ms^(-1)` enters a heavy wooden block and is stopped after a direction of 60cm. The average resistive force exered by the block on the bullet isA. 180NB. 220NC. 270ND. 320N

A `30gm` bullet initially at `120m//s` penetrates `12cm` into a wooden block. The average block. The average resistance exerted by the wooden block is.A. `2850N`B. `2200N`C. `2000N`D. `1800N`

In the previous problem, If the string C is stretched slowly, thenA. The portion AB of the string will breakB. The portion BC of the string will breakC. None of the strings will breakD. None of the above

If a bullet of mass 5gm moving with velocity `100m/sec, penertates the wooden block upto 6cm. Then the average force imposed by the bullet on the block isA. 8300 NB. 417 NC. 830 ND. Zero

A block of metal weighing 2kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1kg/s and at a speed of 5 m/s. Calculate the initial acceleration of the block.