Explore topic-wise InterviewSolutions in .

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.

401.

A train weighing `1000` quintals is running on a level road with a unifrom speed of `72km//h` If the frictional resistance amounts to 50g wt. per quintal find power in watt, take `g = 9.8ms^(2)`.

Answer» Correct Answer - 9800 W
402.

Assertion: On a rainy day, it is difficult to drive a car or bus at high speed. Reason: The value of coefficient of friction is lowered due to wetting of the surface.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is the not correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false.

Answer» Correct Answer - A
On a rainy day, the roads are wet. Wetting of roads lowers the coefficient of friction between the tyres and the road. Therefore, grip on a road of car reduces and thus chances of skidding increases.
403.

A hunter has a machine gun that can fire 50 g bullets with a velocity of `150 m//s` . A 60 kg tiger springs at him with a velocity of `10 ms^(-1)` How many bullets must the hunter fire per secoud into the tiger in order to stop him in his track.

Answer» Here, ` m = 50 g = 50 xx 10^(-3) kg `
` upsilon = 150 m//s , M =60 kg V = 10 m//s `
` n= ` ?
According to the principal of conservation of linear momentum ,
change in momentum of bullets = change in momentum of tiger
` nxx m xx upsilon = MV `
` n = (MV)/(mxxupsilon )=(60xx10)/(50 xx 10^(-3) xx 150 `
` n = 80` (bullet/sec).
404.

A body of mass 40kg resting on a rough horizontal surface is subjected to a force `P` which is just enough to start the motion of the body. If `mu_(s)=0.5mu_(k)=0.4 , g=10ms^(-2)` an dthe force `P` is continuously applied on the body, then the accceleration of the body is.A. zeroB. `1 ms^(-2)`C. `2 ms^(-2)`D. `2.4 ms^(-2)`

Answer» Correct Answer - B
(b) Force, `P=f_(m)=mu_(0) mg`(when body is rest)
When the body starts moving with acceleration a, then `P-F_(k)=ma`
`mu_(0) mg-mu _(k) mg =ma`
`a=(mu_(0)=mu_(k))g`
=(0.5-0.4)10
`=0.1xx10ms^(-2)=1 ms^(-2)`
405.

A body of mass 40kg resting on a rough horizontal surface is subjected to a force `P` which is just enough to start the motion of the body. If `mu_(s)=0.5mu_(k)=0.4 , g=10ms^(-2)` an dthe force `P` is continuously applied on the body, then the accceleration of the body is.A. zeroB. `1m//s^(2)`C. `2m//s^(2)`D. `2.4m//s^(2)`

Answer» Correct Answer - b
When body is just to start moving then force
`F =mu_(s) mg`
When body starts moving with acceleration a then
`F - F_(k) = ma`, where `F_(k) = mu_(k) mg`
or `mu_(s) mg-mu_(k) mg = ma`
or `a= (mu_(s) -mu_(k)) g = (0.5 -0.4) xx 10 = 1ms^(2)` .
406.

A lift is moving down with an acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift, and a man standing stationary on the ground are, respectively.A. `g,g`B. `a,a`C. `(g-a),g`D. `a,g`

Answer» Correct Answer - c
When droped acceleration of the ball is g as will be observed by a man standing stationary on the ground The man inside the lift is having its own downward acc =a Therefore relative acc of the ball as observed by the man in the lift (g-a).
407.

The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator isA. going up and slowing downB. going up and speeding upC. going down and slowing downD. going down and speeding up

Answer» Correct Answer - b,c
If elevator is slowing down acceleration `= -a`
If elevator is speeding up acceleration `=+a`
going up and slowing down
`R - mg = m (-a)`
`R = m (g -a)`
(b) going up and speeding up
`:. R - mg = ma R = m (g + a)`
(c) going down and slowing down
`mg - R = m (-a)` or `R = m (g + a)`
(d) going down and speeding up
`mg - R = ma` or `R = m (g -a)` .
408.

Three blocks `A` , `B` and `C` of masses `4kg` , `2kg` and `1kg` respectively are in contact on a frictionless surface, as shown. If a force of `14N is applied on the `4kg` block, then the contact force between `A` and `B` is. A. 2 NB. 6 NC. 8 ND. 18 N

Answer» Correct Answer - B
409.

The coefficient of static friction `mu_(s)` between block A of mass `2kg` and the table as shown in is 0.1 What would be the maximum mass value of blocks `B` so that the two bloks do not move ? The string and the pulley are assumed to be smooth and massless `(g = 10m//s^(2))` .A. `0.2kg`B. `0.4kg`C. `2.0kg`D. `4.0kg`

Answer» Correct Answer - a
Here, `m_(1) = 2kg, m_(2) = ?`
The two blocks would not move, when
`T = m_(2) g = f_(s) = mu_(s) R = mu_(s) m_(1) g`
`:. M_(2) = mu_(s) m_(1) = 0.1 xx 2 = 0.2kg` .
410.

A `30kg` block rests on a rough horizontal surface A force of `200N` is applied on the block The block acquires a speed of `4m//s` starting from rest in `2s` What is the value of coefficient of friction ? .A. `(10)/(3)`B. `(sqrt(3))/(10)`C. 0.47D. 0.184

Answer» Correct Answer - C
( c) `a=(v)/(t)=(4)/(2)=2ms^(-2) impliesF-f=ma`
or `" "200-muxx30xx10=30xx2`
`therefore" "mu=0.47`
411.

A coin is dropped in a lift. It takes time `t_(1)` to reach the floor when lift is stationary. It takes time `t_(2)` when lift is moving up with costant acceleration. ThenA. `t_(1) = t_(2)`B. `t_(1) gt t_(2)`C. `t_(2) gt t_(1)`D. `t_(1) gt gt t_(2)`

Answer» Correct Answer - B
(b) Time `t_(1)` for stationary lift `=sqrt((2h)/(g))`
When lift is moving up with constant acceleration , then `t_(2)=sqrt((2h)/(g+a))`[Relative acceleration =(g+a)]
`therefore t_(1)gt t_(2)`
412.

In the figure shown, find acceleration of the system and tenslons `T_(1),T_(2) and T_(3) (Take = 10 m//s^(2))`

Answer» `a = ("Net pulling force")/("Total mass")`
`= ((3 + 4 - 2 - 1)g)/(3 + 4 + 2 + 1) = 4m//s^(2)`
`4kg` `40 - T_(1) = 4a :. T_(1) = 24N`
`3kg` `T_(1) + 30 - T_(2) = 3a :. T_(2) = 42N`
`1kg` `T_(3) - 10 = (1) (a) :. T_(3) = 14N`
413.

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider `g=10m//s^2`).A. `16N`B. `20N`C. `22N`D. `4N`

Answer» Correct Answer - b
Let `upsilon` be the velocity given by hand when the ball reaches a height h
`:. (1)/(2)mu upsilon^(2) =mgh`…(i)
If `F` is the force applied by the hand then as per
question `(1)/(2) m upsilon^(2) + F xx 0.2 = mg (h +2)`
Using (i), `mgh + F xx 0.2 = mgh + mg xx 2`
`F = (mg xx 2)/(0.2) = 10 mg = 10 xx 0.2 xx 10 = 20 N` .
414.

Two masses `A` and `B` of mass `15kg` and `6kg` are connected by a string passing over a friction pulley fixed at the corner of a table as shown in The coefficient of friction between `A` and table is `0.3` the minimum mass (in kg) of that must be placed on `A` to prevent it from moving is .

Answer» Correct Answer - `5`
Here, mass of `A =15kg` mass of `B =6kg`
mass of `C = m kg= ?`
To prevent the motion weight of B = force of friction between table and blocks `A` and `C`
`6 g = mu (R) g = 0.3 (15 + m)g`
`15 + m = (6)/(0.3) = 20`
`m = 20 - 15 =5 kg` .
415.

Two masses `A` and `B` of `10 kg` and `5 kg`, respectively , are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown. The coefficient of static friction between `A` and the table is `0.2`. The minimum mass `C` that should be placed on `A` to prevent it from moving is equal to A. 15 kgB. 10 kgC. 5 kgD. 20 kg

Answer» Correct Answer - A
(a) ` mu (m_(A)+m_(C))g=m_(B)g`
`therefore" " m_(C)=(m_(B))/(mu)=m_(A)=(5)/(0.2)-10=15 kg`
416.

Three blocks A,B and C are suspended as shown in the figure. Mass of each block A and C is m. if the system is in equilibrium and mass of B is M, then A. M=2mB. `Mlt2m`C. `Mgt2m`D. `M=m`.

Answer» Correct Answer - b
417.

A boy of `30kg` weight sitting on a horse whips it the horse speeds up at an average acceleration of `2 m//s^(2)` (a) If the boy does not slide back what is the force of friction exerted by the horse on the boy ? (b) If the body slides back during the acceleration what can be said about the coefficient of static friction between the horse and the boy ? Take `g = 10 ms^(2)` .

Answer» (a) As the boy does not slide back , its acceleration = acc of the horce
`:. f_(s) = M a = 30 (2.0) = 60 N`
(b) If the boy slides back the horse could not nexert a friction of `60` N on the boy
Maximum force of static friction `f_(s) = mu_(s) R= mu_(s) (M g) :. mu_(s) (30) (10) le 60 N`
`:.` Max value of coeff of static friction `= mu_(s) = (60)/(300) = 0.20` .
418.

Reading shown in two spring balances `S_(1)` and `S_(2)` is 90kg and 30kg respectively and lft is accelerating upwards with acceleration `10m//s^(2).S_(1)` in elongated and `S_(2)` is compressed. The mass is stationary with respect to life. Then the mass of the block will be A. 60kgB. 30kgC. 120kgD. None of these

Answer» Correct Answer - a
419.

Earth is a rotating frame of reference, even then it is considered as inertial frame of reference for all practical purposes. Why ?

Answer» Earth is revoling around the sun. Thereforce , it possesses centripetal acceleration . Therefore , it should be treated as non inertial frame of reference. However. Value of centripetal acceleration of earth ,
` a = (upsilon^(2))/R = omega^(2) R = ((2 pi)/(T))^(2) R = (4pi^(2)R)/(T^(2)) `
Putting ` R = 1 A .U = 1 .496 xx 10^(11) m , `
and ` T = 365 (1)/(4)` days
`= (1461)/(4) xx 24 xx 60 xx 60 xx sec`
` = 3.16 xx 10^(7) sec`, we get
`a = 4 xx (22)/(7) xx (22)/(7) xx (1.496 xx 10^(11))/((3.16 xx 10^(7) )^(2))`
` = 0.006 m//s^(2)`
which is negligibly small compared to acceleration due to gravity `(g = 9.8 m//s^(2))` hence for terrestrial experiments , earth can be considered as an inertial frame .
420.

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity `omega` is an example of non=inertial frame of reference. The relationship between the force `vecF_(rot)` experienced by a particle of mass m moving on the rotating disc and the force `vecF_(in)` experienced by the particle in an inertial frame of reference is `vecF_(rot)=vecF_(i n)+2m(vecv_(rot)xxvec omega)+m(vec omegaxx vec r)xxvec omega`. where `vecv_(rot)` is the velocity of the particle in the rotating frame of reference and `vecr` is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter fo a disc of radius R rotating counter-clockwise with a constant angular speed `omega` about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis `(vecomega=omegahatk)`. A small block of mass m is gently placed in the slot at `vecr(R//2)hati` at `t=0` and is constrained to move only along the slot. The net reaction of the disc on the block isA. (a) `1/2momega^2R(e^(2omegat)-e^(-2omegat))hatj+mghatk`B. (b) `1/2momega^2R(e^(omegat)-e^(-omegat))hatj+mghatk`C. (c) `-momega^2Rcosomegathatj-mghatk`D. (d) `momega^2Rsinomegahatj-mghatk`

Answer» Correct Answer - B
`vecF_(rot)=vecF_(i n)+2m(V_(rot)hati)xxwhatk+m(whatkxxrhati)xxwhatk`
`:.` `mromega^2hati=vecF_(i n)+2mV_( rot) omega(-hatj)+momega^2rhati`
`vecF_(i n)=mrV_(rot)omegahatj` -(i)
But `r=(R)/(4)[e^(wt)+e^(-wt)]`
`:.` `(dr)/(dt)=V_r=(R)/(4)[omegae^(wt)-omegae^(-wt)]` -(ii)
From (i) and (ii)
`vecF_(i n)=2m(Romega)/(4)(e^(wt)-e^(-wt))omegahatj`
`:.` `vecF_(i n)=(mRomega^2)/(2)(e^(wt)-e^(-wt))hatj`
`:.` `vec F_(reaction)=(mRomega^2)/(2)(e^(wt)-e^(-wt))hatj+mghatK`
421.

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity `omega` is an example of non=inertial frame of reference. The relationship between the force `vecF_(rot)` experienced by a particle of mass m moving on the rotating disc and the force `vecF_(in)` experienced by the particle in an inertial frame of reference is `vecF_(rot)=vecF_(i n)+2m(vecv_(rot)xxvec omega)+m(vec omegaxx vec r)xxvec omega`. where `vecv_(rot)` is the velocity of the particle in the rotating frame of reference and `vecr` is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter fo a disc of radius R rotating counter-clockwise with a constant angular speed `omega` about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis `(vecomega=omegahatk)`. A small block of mass m is gently placed in the slot at `vecr(R//2)hati` at `t=0` and is constrained to move only along the slot. The distance r of the block at time isA. (a) `R/4(e^(omegat)+e^(-omegat))`B. (b) `R/2cos omega t`C. (c) `R/4(e^(2omegat)+e^(-2omegat))`D. (d) `R/2 cos 2omegat`

Answer» Correct Answer - A
Force on the block along slat `=mr omega^2=mv(dv)/(dr)`
`:.` `int_o^vVdv=int_(R//2)^romega^2rdrimpliesV=omegasqrt(r^2-(R^2)/(4))=(dr)/(dt)`
`:.` `int_(R//4)^r(dr)/(sqrt(r^2-(R^2)/(4)))=int_o^t omegadt`
On solving we get
`r+sqrt(r^2-(R^2)/(4))=(R)/(2)e^(wt)`
or `r^2-(R^2)/(4)=(R^2)/(4)e^(2wt)+r^2-2r(R)/(2)e^(wt)`
`:.` `r=(R)/(4)(e^(wt)+e^(-wt))`
422.

One aften comes across the following kind of statement concerning circular motion A particle moving uniformly along a circle experiences a force directed towards the center and an equal and opposite force directed away from the centre The two forces together keep the particle in equilibrium Explain what is wrong with the statement ?

Answer» The statement is wrong because an external force is required to move a body uniformaly in a circle This is centripetal force which acts on the particle .
The centrifugal force which is equal and opposite to the centripetal force comes into play on account of tendency of the particle to get back to the straight line path This force does not act on the straight line path This force does not act on the particle It acts on the agent supplying the necessary centripetal force .
423.

A block is over a plank . The coefficient of friction between the block and the plank id `mu=0.2 `Initially both are at rest , suddenly the plank starts moving with accleration `alpha_(0)=4m//s^(2)` The displacement of the block in s is `(g=10m//s^(2))`A. 1m relative to groundB. 1 m relative to plankC. zero relative to plankD. 2 m relative to ground

Answer» Correct Answer - a,b
424.

A rod of length `l(lt2R0` is kept inside a smooth sperical shell as shwon in fifure .mass of the rod is m The normal reaction when l=R isA. `(mg)/(2)`B. `(mg)/(4)`C. `(mg)/(2sqrt(3))`D. `(mg)/(sqrt(3))`

Answer» Correct Answer - d
425.

A rod of length `l(lt2R)` is kept inside a smooth sperical shell as shwon in fifure .mass of the rod is m Keeping mass to be constant I flength of the rod is increased `("but always " lt 2R )` the normal reaction at two ends of the rod A. will remain constantB. will increaseC. will decreaseD. may increase or decrease

Answer» Correct Answer - b
426.

In the pulley system shown in fifure the movable pullet=ys A,B and C are of mass 1 kg each D and E are fixed pulleys are light and inexensible. Choose the correct alternative (s).A. Tension in the string is 6.5 NB. Acceleration of pulley A is g/3 downwardC. Acceleration of pulley B is g.6 upwardD. Acceleration of pulley C is g/3 upward

Answer» Correct Answer - a,b,d
427.

In the arrangement shown in figure all surface are amoth .Select the correct alternative(s). A. For any valure of `theta` acceleration of A and B are equalB. Contact force between the two blocks is zero if `m_(A)//m_(B)=tantheta`C. Contact force between the two is zero for any value of `m_(A) or m_(B)`D. Normal reactions by the wedge on the block are equal

Answer» Correct Answer - a,c
428.

A gramphone record is revolving with an angular velocity `omega`. A coin is placed at a distance `R` from the centre of the record. The static coefficient of friction is `mu`. The coin will revolve with the record ifA. `r =mugomega^(2)`B. `r lt(omega^(2))/(mug)`C. `rlt(mug)/(omega^(2)`D. `r ge(mug)/(omega^(2))`

Answer» Correct Answer - c
The coin will revolve with the record if centrifugal
force `MR omega^(2) lt f` (force of friction)
`mr omega^(2) lt muR`
or `mromega^(2) lt mu mg`
`r lt (mug)/(omega^(2))` .
429.

A bend in a level road has a radius of 100 m Find the maximum speed which a car turning this bend may have without skidding if coefficient of friction between the tyres and the road is 0 .8 S .

Answer» Correct Answer - `28m//s` .
Here, `r = 100 m, upsilon_(max) = ? Mu = 0.8`
The car will not skid if force of friction provides the necessary centripetal force
`f = mu = (m upsilon_(max)^(2))/(r) `
`:. upsilon_(max) = sqrt(mu rg) = sqrt(0.8 xx 100 xx 9.8) =28m//s` .
430.

In the arrangement shown in figure wedge of mass M moves towards left with an acceleration a. All surfaces are smooth. The acceleration of mass m relative to wedge is A. `a//2`B. `(2Ma)/(m)`C. `(2(M+m)a)/(m)`D. `((M+m)a)/(m)`

Answer» Correct Answer - C
431.

Starting from rest a body slides down a `45^(@)` inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of the body and the inclined plane is :A. 0.2B. 0.25C. 0.75D. 0.5

Answer» Correct Answer - C
( c) `s=(1)/(2)at^(2) " or " t=sqrt((2s)/(a)) " or " t prop (1)/(sqrt(a))`
`(t_(1))/(t_(2))=sqrt((a_(2))/(a_(1)))`
or `" " 2=sqrt((g sin theta)/(g sin theta-mug cos theta))=sqrt((1)/(1-mu))`
as`" " sin 45^(@)=cos 45^(@)`
`therefore" " 1- mu =(1)/(4) implies mu=0.75`
432.

Consider the shown arrangement. Assume all surfaces to be smooth. If N represents magnitudes of normal reaction between block and wedge, then acceleration of M along horizontal is equla to A. `(N sin theta)/(M)` along + veX-axisB. `(N cos theta)/(M)` along - veX-axisC. `(N sin theta)/(M)` along - veX-axisD. `(N sin theta)/(m+M)` along - veX-axis

Answer» Correct Answer - C
( c) Wedge move due to horizontal component of normal reaction.
Thus `" " a=(N_(H))/(M)=(N sin theta)/(M)" " `(along-ve X-axis)
433.

A body of mass `50kg` is put on a spring weighing machine in a lift What is the reading of the machine when (i) Lift ascends with a unifrom velocity of `10m//s` (ii)descends with an acceleration of `5m//s^(2)` ? Take `of = 10m//s^(2)` .

Answer» Correct Answer - `50kg f; 25kg f` .
Here, `m = 50 kg. R_(1) = ? upsilon = 10 m//s a_(1) = 0`
`R_(2) = ? A_(2) = 5 m//s^(2) g =10 m//s^(2)`
`R_(1) = m (g +a_(1)) = 50 (10 + 0) = 500 N = 50 kg f`
`R_(2) = m (g - a_(2)) = 50 (10 -5) = 250 N = 25 kg f` .
434.

A stone of mass `5kg` falls from the top of a cliff `50m` high and buries `1m` deep in sand . Find the average resistance offered by the sand and the time it takes to penetrate .

Answer» Correct Answer - 2450 N, `0.064s`
435.

In the adjoining figure if acceleration of M with respect to ground is a then A. acceleration of m with respect to M is aB. acceleration of m with respect to ground is 2 a sin `(alpha/2)`C. acceleration of m with respect to fground is aD. acceleration of mawith respect to ground is a tan `alpha`

Answer» Correct Answer - a,b
436.

A rocket consumes `24kg` of fuel per second The burnt gases escape the rocket at a speed of `6.4km//s` relative to the rocket Calculate the upthrust recevied by the rocket Also calculate the velocity acquired by the rocket when the mass redues to `1//100th` of its initial mass.

Answer» Correct Answer - `1.536 xx10^(5) N; 2.984 xx10^(4) m//s` .
Here,` (dm)/(dt) = 24 kg//s`
`u = 6.4 km//s = 6.4 xx 10^(3) m//s`
upthrust ` = u ((dm)/(dt)) = 6.4 xx 10^(3) xx24`
`=1.536 xx 10^(5)N`
As `upsilon = u log_(e) (m_(0)/(m))`
`:. Upsilon = 6.4 xx 10^(3) log_(e) ((m_(0))/(m_(0) //100))`
`= 64 xx 10^(3) xx 2.303 log_(10) (100)`
`upsilon = 6.4 xx 10^(3) xx 2.303 xx 2`
`2.948 xx 10^(4) m//s` .
437.

In the figure, `m_(A)=2 kg` and `m_(B)=4kg`. For what minimum value of F,A starts slipping over B? `(g=10m//s^(2))` A. 24NB. 36NC. 12ND. 20N

Answer» Correct Answer - b
438.

In the figure shown if coefficient of friction is `mu`, then `m_(2)` will start moving upwards if A. `(m_(1))/(m_(2)) gt sin theta - mu cos theta`B. `(m_(1))/(m_(2)) gt sin theta + mu cos theta`C. `(m_(1))/(m_(2)) gt mu sin theta - cos theta`D. `(m_(1))/(m_(2)) gt mu sin theta + cos theta`

Answer» Correct Answer - B
(b) `m_(1)g gt m_(2)g sin theta+mum_(2)g cos theta`
`implies" " (m_(1))/(m_(2))gt sin theta+ mu cos theta`
439.

A horizontal force F is applied on a riing of mass `m_(1) ` constrained to move on a horizontal smooth wire the hanging mass `m_(2)` is constant angle with A. Force on `m_(1)` by wire is less than `(m_(1)+ m_(2))g`B. Net force on `m_(2) ` is `(m_(2) F)/(m_(1)+m_(2))`C. Tension in rod is `m_(2)g sectheta`D. `F=(m_(1)+m_(2)) g tan theta`

Answer» Correct Answer - b,c,d
440.

Two mankeys (a) and (b ) of same mass m=1 kg are hanging on the dtring such that block of amss 2 kg ramains at rest and it is given that monkeys( B) is just holding the string then which of the following statement (S) is /are correct `(g=10m//s^(2))` A. Acceleration of maney (B ) is `10m//s^(2)` upwardsB. Acceleration of mankey (A) is `30m//s^(2)` upwardsC. Acceleration of mankey (A) with respect to his rope `35 m//s^(2)` upwards.D. Acceleration on mankey ( B) with respect to his rope is `50m//s^(2)` upwards.

Answer» Correct Answer - a,b,c
441.

Two unequal masses are connected on two sides of a light string passing over a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped for a moment, 1.0 s after the system is set into motion. The time elapsed before the string is tight again is `(g=10 m//s^(2))` A. `1//4s`B. `1//2 s`C. `2//3 s`D. `1//3 s`

Answer» Correct Answer - D
442.

A body is dropped from the celling of a transparent cabin falling freely towards the earth . Describe the motion of the body as observed by an observer (a) sitting in the cabin (b) standing on earth.

Answer» (a) The body will appear stationary in air , (b) The body will appear to be falling freely under acceleration due to gravity .
443.

A hammer of mass 1 kg moving with speed of `6 ms^(-1)` strikes a wall and comes to rest in 0.1 s . Calculate (i) the impules of force (ii) the retardation of the hammer , and (iii) the retarding that stops the hammer .

Answer» (i) Impulse =`Fxxt`=m(v-u)=1(0-6)=-6 Ns
(ii) Average retarding force that stops the hammer
`F=("Impulse")/("Time")=(6)/(0.1)`=60 N
(iii) Average retardation, `a=(F)/(m)=(60)/(1)=60ms^(-2)`
444.

Four blocks of the same mass m connected by cords are pulled by a force F on a smooth horizontal surface as shown in Determine the tensions `T_1 ,T_2` and `T_3` in the cords.

Answer» Correct Answer - `T_(1)=(3)/(4)F,T_(2)=(1)/(2)F, T_(3)=(1)/(4)F`
445.

As shown in Fig. three masses m, 3m and 5m connected together lie on a frictionless horizontal surface and pulled to the left by a force F. The tension `T_(1)` in the first string is 24 N. Find (i) acceleration of the system, (ii) tension in the second string, and (iii) force F.

Answer» Correct Answer - (i) a=3/m (ii) `T_(2)=15N` (iii) F = 27
446.

A wooden block of mass 100 kg rests on a flat wooden floor, the coefficient of friction between the two being `0.4`. The block by a rope making an angle of `30^(@)` with the horizontal. What is the minimum tension along the rope that just makes the rope sliding?

Answer» Correct Answer - `367.7N`
447.

A truck moving at `72 kmh ^(-1)` carries a steel girder which rests on its wooden floor. What is the minimum time in which the truck can come to stop without the girder moving forward ? Coefficient of static friction between steel and wood is `0.5`.

Answer» Correct Answer - `4.08s`
448.

A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

Answer» Correct Answer - `9.8 N`
As the block is at rest therefore maximum force of static friction
`f_(s) = mu _(s) R` But `R = m g cos theta`
`:. f_(s) = mu_(s) m g cos theta = 0.7 xx 2xx 9.8 cos 30^(@)`
`= 1.4 xx 9.8 xx 0.866 = 11.9 N`
But in this case actual force of static friction cannot exceed the force due to gravity `(mg sin theta)` acting down the plane
`:.` Actual force of static friction
`= mg sin theta = 2 xx 9.8 xx sin 30^(@) = 9.8 N`.
449.

A block of mass `10kg` is placed on a rough horizontal surface having coefficient of friction `mu=0.5` . If a horizontal force of `100N` is acting on it, then acceleration of the will be.A. zeroB. `10 ms^(-2)`C. `5 ms^(-2)`D. `5.2 ms^(-2)`

Answer» Correct Answer - C
( c) `a=(F-mu mg)/(m)=(100-0.5xx10xx10)/(10)=5ms^(-2)`
450.

A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isA. 9.8 NB. `0.7xx9.8xxsqrt(3) N`C. `9.8xxsqrt(3) N`D. `0.7xx9.8 N`

Answer» Correct Answer - A
(a) Angle of respose `theta_(r)=tan ^(-1)(mu_(s))=tan^(-1)(0.7)`
or `tan theta_(r)=0.7`
Angle of plane is `theta=30^(@), tan theta= tan 30^(@)=0.577`
Since`" " tan theta lt tan theta_(r), theta lt theta_(r)`
Block will not slide or `f= mg sin theta ne mu mg cos theta`
or `" " f=(2)(9.8) sin 30^(@)=9.8 N`