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351.

What is the angle of friction between two surfaces in contact , if coefficient of friction is `1//sqrt(3)` ?

Answer» ` mu = (1)/sqrt3 , theta = ` ?
As ` tan theta = mu = (1)/sqrt3 :. theta = 30^(@)` .
352.

Can we get off a frictionless horizontal surface by umppin ?

Answer» No as frictionless surface offers no reaction .
353.

A particle stays at rest as seen in a frame. We can conclude thatA. the frame is inertialB. resultant force on the particle is zeroC. the frame may be inertical but resultant force on the particle is zeroD. the frame may be non-inertial but there is a non zero resultant force

Answer» Correct Answer - c,d
particle will be seen at rest only when frame is inertial and resultant force on particle is zero Also if frame is non inertial (accelerated) the particle must also possess the same acc in magnitude and direcction resultant force on the particle must be non zero .
354.

A girl press her physics text book against a rough vertical wall with her hand. The direction of the frictional force on the book exerted by the wall isA. downwardsB. upwardsC. out from the wallD. into the wall

Answer» Correct Answer - B
Friction forces are always parallel to the surfaces in contact, while in this case, are the wall and the cover of the book. This tells us that the friction force is either upwards or downwards. Because the tendency of the book is to fall due to gravity, the friction force must in the upward direction.
355.

What is the angle between frictional force and instantaneous velocity of the body moving over a rough surface ?

Answer» The angle is `180^(@)` , because force of fricting always opposes the relative motion .
356.

Is frictions a self adjusting force ?

Answer» No all types of friction are not self adjusting Only stactic friction is a self adjusting force .
357.

Which of the following is a self adjusting force ?A. Static frictionB. Rolling frictionC. Sliding frictionD. Dynamic friction

Answer» Correct Answer - A
Static friction is a self adjusting force.
358.

How does a lubricant help in reducing friction ?

Answer» When a lubricant is added to a machine it spreads between the two surface rubbing each other fills the irreularities present on the surfaces and forms a thin layer between the surfaces in contact As a result of it the contact between the two hard surfaces is replaced by the contact between the hard surface and lubricant layer Due to it the force of friction is reduced considerably .
359.

Which one of the following can also act as a lubricant in the machines?A. Iron fillingsB. Polish on machinesC. Flow of waler through the machineD. Flow of compressed and purifie air.

Answer» Correct Answer - D
In the machines, the flow of compressed and purified air lowers the friction, hence is acts as a lubricant.
360.

In the previous problem `3` the magnitude of the momentum transferred during the hit is .A. zeroB. `0.75"kgms"^(-1)`C. `1.5"kgms"^(-1)`D. `14"kgms"^(-1)`

Answer» Correct Answer - C
The magnitude of the momentum transferred during the hit is
`|Deltabar(p)|=sqrt((-0.9)^(2)+(-1.2)^(2))kgms^(-1)`
`=sqrt(0.81+1.44)kgms^(-1)=1.5 kgms^(-1)`
361.

In the previous problem `3` the magnitude of the momentum transferred during the hit is .A. zeroB. `0.75 kgms^(-1)`C. `1.5kgms^(-1)`D. `14kg ms^(-1)`

Answer» Correct Answer - c
Magnitude of momentum transferred (=change in momentum) `|vec(p)| = sqrt(0.9^(2) + (1.2)^(2)) = 1.5kgms^(-1)` .
362.

A 1.5 kg block placed on a frictless orizontal floor is pulled by a force F Direction of the force is unchanged and its magnitude is incrcresaed with time according to the equation F =0.5 t n where t denotes time in second . Which of the following is /are correct for subsequunt motion after `t=0` s? [`g=10m//s^(2)`] A. The block leaves the flor at the instant t=50s.B. Acceleration of the block at the innstant t=40s. `8m//s^(2)` towards the right .C. Normal reaction from the ground on the block at the instant t=50 s is 8 N.D. Acceleration vector of the block at the instant t=80s. Is `(1.3hati +6hatj ) m//s^(2)`.

Answer» Correct Answer - a,d
363.

A body rolled on ice with a velocity of `8 ms^(-1)` comes to rest after travelling a distance of 4 m . Calculate the coefficient of friction .

Answer» Here , ` u = 8 m // s , upsilon = 0 , s = 4 m , a = ? `
From ` v^(2) - u^(2) = 2` as
`0 - 8^(2) = 2 (a) 4 = 8 a`
` a = - (64)/(8) = - 8 m// s^(2) `
Negative sigh represents retardation of body
` mu = (f)/(R) = (ma)/(mg) = (a)/(g) = (8)/(9.8) = 0.816` .
364.

A cricket ball is rolled on ice with a velocity of `5.6 m//s` and comes to rest after travelling 8 m . Find the coefficient of friction Given `g = 9.8 m//s^(2)` .

Answer» Here , ` u = 5.6 m//s , upsilon = 0 , s = 8 m , g = 9.8 m//s^(2) `
From ` upsilon^(2) - u^(2) = 2 a s `
` 0 - (5.6)^(2) = 2 xx a xx 8 `
`:. a = (5.6 xx 5.6)/(16 )=- 1.96 m//s^(2) `
` mu_(k) = (a)/(g) = (1.96)/(9.8)=0 .2 ` .
365.

Conside the situation shown in figure. The block `B` moves on a frictionless surface ,while the coefficient of friction between `A` and the surface on which it moves is `0.2`. Find the acceleration with which the masses move and also the tension in the strings. `(Take g = 10 m//s^(2))`.

Answer» Let a be the acceleration with which the masses move and `T_(1)` and `T_(2)` be the tensions in left and right strings. Friction on mass `A` is `mumg =8N`. Then equationof motion `A,B` and `C` are
For mass `A` `T_(1) -8 = 4a` …(i)
For mass `B` `T_(2) = 8a` …(ii)
For mass `C` `200-T_(1) -T_(2) = 20a` …(iii)
Adding the above three equation, we get `32a =192`
or `a = 6m//s^(2)`
From Eqs,(i) and (ii), we have `T_(2) = 48N`
and `T_(1) =32N`
366.

A nucleus is at rest in the laboratory frame of reference Show that if it distintegrates into two smaller nuclei the products must be emitted in opposite directions .

Answer» Let `m_(1) m_(2)` be the masses of products and `vec(upsilon_(1)),vec (upsilon_(1))` be thir respective velocities Therefore total linear momentum after disintegration ` = m_(1) vec(upsilon_(1)) + m_(2) vec (upsilon_(2)) ` Before disintegration the nucleus is at rest Therefore its linear momentum before disintegration is zero
According to the principle of conservation of linear momentum
` = m_(1) vec(upsilon_(1)) + m_(2) vec(upsilon_(2)) = 0 or vec(upsilon_(2)) = (-m_(1) vec(upsilon_(1)))/m_(2) `
Negative sign shown that `vec(v_(1)) and vec(v_(2)` are in opposite directions.
367.

A uniform rod of length L and density `rho` is being pulled along a smooth floor with a horizontal acceleration `alpha` (see Fig.) The magnitude of the stress at the transverse cross-section through the mid-point of the rod is……..

Answer» Correct Answer - A::B
Let A be the area of cross-section of the rod.
Consider the back half portion of the rod.
Mass of half portion of the rod `=(rhoAL)/(2)`
The force responsible for its acceleration is
`f=(rhoAL)/(2)xxa` `:.` Stress `=f/A=(rhoLalpha)/(2)`
368.

In the arrangement shown in figure `m_(1)=1kg, m_(2)=2kg`, pullyes are massless and strings are light. For what value of M the mass `m_(1)` moves with constant velocity? (Neglect friction) A. 6kgB. 4kgC. 8kgD. 10kg

Answer» Correct Answer - c
369.

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by A. `sqrt2mg`B. `sqrt2mg`C. `[sqrt((M + m)^(2) + m^(2))]g`D. `[sqrt((M + m)^(2) + m^(2))]g`

Answer» Correct Answer - c
Force on the pulley by the clamp = resultant of `T = (M + m)` g and mg acting along horizontal and vertical directions respectively
`= sqrt([(M + m) g]^(2) + (mg)^(2))`
`=[sqrt((M + m)^(2) + m^(2)]]g` .
370.

A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with unifrom acceleration the force exerted on the car is .A. `(mv)/(2)` eastward and is exerted by the car engine.B. `(mv)/(2)` eastward and is due to the friction on the tyres exerted by the road.C. more than `(mv)/(2)` eastward exerted due to the 2 engine and overcomes the friction of the road.D. `(mv)/(2)` exerted by the engine.

Answer» Correct Answer - B
Here, Mass of the car=m
Initial velocity, u=0 (As the car starts from rest)
Final velocity `bar(v)=vbar(i)` along east
t=2s
Using, `v=u+atRightarrow v hati=0+bar(a)xx2 or bar(a)=(v)/(2)hati`
Force exerted on the car is `bar(F)=mbar(a)=(mv)/(2) hat(i)=(mv)/(2)"eastward"`
This is due to the friction on the tyres exerted by the road.
371.

A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with uniform acceleration the force exerted on the car is .A. `(mv)/(2)` eastward and is exerted by the car engineB. `(mv)/(2)` eastward and is due to the friction on the tyres exerted by the roadC. more than `(mv)/(2)` eastward exerted due to the engine and overcomes the friction of the roadD. `(mv)/(2)` exerted by the engine

Answer» Correct Answer - B
Given , mass of the car = m
As car starts from rest u = 0
Velocity acquired along east = `v hati`
Duration = t = 2s
We know that `" " v = u + at`
`implies " " vhati = o + a xx 2`
`implies " " a = (v)/(2) hati`
Force , `" " F = ma = (mv)/(2) hati`
Hence , force acting on the car is `(mv)/(2)` towards east . As external force on the system is only friction hence , the force `(mv)/(2)` is by friction . hence force , by engine is internal force .
372.

A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with unifrom acceleration the force exerted on the car is .A. `(m upsilon)/(2)` eastward and is exerted by the car engineB. `(m upsilon)/(2)` eastward and is due to the friction on the tyres exerted by the roadC. more than `(m upsilon)/(2)` eastward exerted due to the engine and overcomes the friction of the roadD. `m (upsilon)/(2)` exerted by the engine

Answer» Correct Answer - b
Here, mass of car `=m`
As it starts from rest, `u = 0` final velocity along east `upsilon = upsilonhati , t =2 sec`
From `upsilon = u + at`
`upsilonhati = 0 + vec (a) xx 2, vec(a) = (upsilon)/(2) hati`
`vec(F) = m vec(a) =( m upsilon)/(2) hati` ,force of car is` (m upsilon)/(2)` east ward
This is due to friction on the tyres exerted by the road .
373.

A cricket ball of mass 150 g has an intial velocity u = `(3 hati + 4 hatj)ms^(-1)` and a final velocity `v = -(3hati + 4 hatj) ms^(-1)` , after being hit. The change in momentum (final momentum - initial momentum ) is (in `Kg ms^(1)`)A. zeroB. `-(0.45 hati+0.6 hatj)`C. `-(0.9 hati+1.2 hat j)`D. `-5( hati+hatj) hati`

Answer» Correct Answer - C
( c) Given,`u=(3 hati+4 hatj)ms^(-1) " and " v=-(3hati+4hatj) ms^(-1)`
Mass of the ball=150g=0.15 kg
`Deltap`=change in momentum
= Final momentum-Initial momentum
=mv-mu
`=m(v-u)=(0.15)[-(3 hati +4 hatj)-(3 hati+4 hatj)]`
`=(0.15)[-6hati-8hatj)]=-]0.15xx6hati+0.15 xx 8hatj)]`
`=-[0.9 hati+1.2 hatj]`
Hence, `Deltap=-[0.9 hati+1.2 hatj]`
374.

The velocity- time graph of the figure shown the motion of a wooden block of mass `1 kg` which is given an intial push at `t = 0` along a horizontal table. A. The coefficient of friction between the block and the table is `0.1`B. The coefficient of friction between the block and the table is `0.2`C. If the table was half of its present roughness, the time taken by the book to complete the journey is `4 s`D. If the table was half of its present roughness, the time taken by the book to complete the journey is `8 s`

Answer» Correct Answer - A::D
`a = "slop of" v = t "graph"`
`= - 1 m//s^(2)`
`:. "Retardation" = 1 m//s^(2) = (mu mg)/(m) = mu g`
or `mu = (1)/(g) = (1)/(10) = 0.1`
If `mu`is half ,then reladation a is also half .So using
`v =u - at`
or `0 = u - at`
or `t = (u)/(a)` or `t prop (1)/(a)`
we can see that `t` will be two times.
375.

A ball moving with velocity `2 ms^(-1)` collides head on with another stationary ball of double the mass. If the coefficient of restitution is `0.5`, then their velocities (in `ms^(-1)`) after collision will beA. 0,1B. 1,1,C. 1,0,5D. 0,2

Answer» Correct Answer - A
376.

A large force is acting on a body for a short time. The impulse imparted is equal to the change inA. accelerationB. momentumC. energyD. velocity

Answer» Correct Answer - B
If a large force F acts for a short time dt the impulse imparted I is
`I=Fdt=(dp)/(dt)dt,I=dp="change in momentum"`
377.

A woman throws an object of mass `500g` with a speed of `25 ms^(-1)` . (a) What is the impulse imparted to the object ? (b) If the object hits a wall and rebounds with half the original speed what is the change in momentum of the object ?

Answer» Here, ` m = 500 g = (1)/(2) kg , u = 25 ms^(-1) `
(a) Impulse imparted = change in momentum = initial momentum given ` m u = (1)/(2) xx25 = 12.5 Ns `
(b) On rebounding force the wall ` upsilon = -(1)/(2) xx 25 = - 12.5 m//s `
Change in momentum ` = m (upsilon- u ) = (1)/(2) (-12.5- 25) = - 18.75 kg ms^(-1)` .
378.

A particle of mass 0.2 kg attached to a massless string in a verticle circle of radius 1.2 m. It is imparted a speed of `8 ms^(-1)` at the lowest point of its circular path. Does the particle complete the verticle circle ? What is the change in tension in the string when the particle moves from the position where the string is vertical to the position where the string is horizontal ?

Answer» Here, ` m = 0.2kg , r =1.2 m , upsilon_(L) = 8 ms^(-1)`
Minimum speed at lowest point for completing the verticle circle ` =sqrt(5 g r) = sqrt(5 xx 10 xx 1.2) = 7.75 m^(-1) `
As ` upsilon = 8m//s ` is greater than` 7.75ms^(-1)`, therefore , the particle completes the vertical circle .
From ` T = (m upsilon^(2))/(r) + mg cos theta`
At the lowest point , string is vertical
` T_(L) = (m upsilon_(L)^(2))/(r) + mg cos 0^(@) = (m upsilon_(L)^(2))/(r) + mg `
When the string is horizontal say at A
` T_(A) = (m upsilon_(A)^(2))/(r) + mg cos90^(@) = (m upsilon_A^(2))/(r) `
` T_(L) - T_(A) = (m)/(r) (upsilon_(L)^(2) - upsilon_(A)^(2)) + mg = (m)/(r) xx (2 g r ) + mg = 3 mg = 3 xx 0.2 xx 10 = 6 N ` .
379.

A lift of mass 100 kg is moving upwards with an acceleration of 1 `m//s^(2)`. The tension developed in the string, which is connected to lift is (`g=9.8m//s^(2)`)A. 980 NB. 1080 NC. 1100 ND. 1000 N

Answer» Correct Answer - B
380.

The upper half of an inclined plane with inclination `phi` is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given byA. `mu=2 tan theta`B. `mu= tan theta`C. `mu=(2)/(tan theta)`D. `mu=(1)/(tan theta)`

Answer» Correct Answer - A
(a) `v^(2)=2a_(1)s(a_(1)=g sin theta)`
`0=v^(2)-2a_(2)s(a^(2)=mu g cos theta-g sin theta)`
From these two equations we see that `a_(1)=a_(2)`
`implies" " g sin theta=mu g cos theta-g sin theta implies 2 tan theta =mu`
381.

The upper half of an inclined plane with inclination `phi` is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given byA. (a) `2 cos phi`B. (b) `2 sin phi`C. (c) `tan phi`D. (d) `2 tan phi`

Answer» Correct Answer - D
According to work-energy theorem, `W=DeltaK=0`
(Since initial and final speeds are zero)
`:.` Work done by friction + Work done by gravity =0
i.e., `-(mu mg cos phi)l/2+mg l sin phi =0`
or `mu/2 cos phi = sin phi` or `mu=2 tan phi`
382.

A balloon starting from rest ascends vertically with uniform acceleration to a height of 100m is 10s. The force on the bottom of the balloon by a mass of 50 kg us (`g=ms^(-2)`)A. 100 NB. 300 NC. 600 ND. 400 N

Answer» Correct Answer - C
( c) Given that, h=100m and t=10s
From equation of motion,
`h=(1)/(2) at^(2)`
`a=(2h)/(t^(2))=(2xx100)/((10)^(2))=2 ms^(-2)`
Now, `F=50(10+2) implies F=600 N`
383.

A stream of water flowing horizontal with a speed of `15 ms^(-1)` gushes out of tube of cross sectional area `10^(2) m^(2)` and hits at a vertical wall nearby . What is the force exerted by the impact of water, assuming that water rebounds with the same speed.

Answer» Correct Answer - `4500N` .
Here, `upsilon =15m//s, A = 10^(-2) m^(2)`
Volume of water gushing out//sec
`=A xx upsilon = 10^(-2) xx 15m^(3)//s`
Mass of water flowing out/sec = Volume `xx` density
`m = 15 xx 10^(-2) xx 10^(3) = 150kg//s`
As water rebounds with the same speed, therefore Impact of water = force on the wall
= change in linear momentum/sec
=mass/sec `xx` change in velocity
`=150 xx (15 + 15) = 4500N`.
384.

A helicopter of mass 1000 kg rises with a vertical acceleration of `15 ms^(-2)`. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the (a) force on the floor by the crew and passengers, (b) action of the rotor of the helicopter on the surroundings air, (c) force on the helicopter due to the surroundings air.

Answer» Here ,mass of helicopter , `m_1` = 1000 kg .
Mass of the crew and passengers , `m_2` = 300 kg upward acceleration , a = 15` ms^(-2)` and g = 10`ms^(-2)`
(a) Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers = `m_2(g + a) ` = 300 (10 +15) N = 7500 N
(b) Action of rotor of helicopter on surrounding air is obviously vertically downwards , because helicopter rises on account of reaction to this force. Thus , force of action
F = `(m_1 + m_2)`(g + a) = ( 1000 + 300) (10 +15) = `1300 xx 25 ` = 32500 N
(c ) Force on the heliopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore force of reaction , F = 32500 N , verically upwards.
385.

A helicopter of mass 2000 kg rises with a vertical acceleration of `15ms^-2`. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (g`=10ms^-2`) (a) Force on the floor of the helicopter by the crew and passengers. (b) action of the rotor of the helicopter on the surrounding air. (c ) force on the helicopter dur to the surrounding air.

Answer» Given mass of helicopter `(m_(1))` = 2000 kg
Mass of the crew and passengeres `m_(2) = 500` kg
Acceleration in vertical direction a = 15 m/ `s^(2) (uarr)` and g = 10 m/ `s^(2) (darr)`
(a) Force on the floor of the helicopter by the crew and passengers
`m_(2)(g +a ) = 500 (10 + 15) N`
`500 xx 25 N = 12500 N`
(b) Acting of the rotor of the helicopter on the surronding air = `(m_(1) + m_(2)) (g + a)`
= `(2000 + 500) xx (10 + 15) = 2500 xx 25`
= `625000 N` (downward)
(c) Force on the helicopter due to the surrounding air
= reaction of force applied by helicopter .
= 62500 N (upward)
386.

A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is A. always radially outwardsB. always radially inwardsC. radially outwards initially and radially inward laterD. radially inwards initially and radially outwards later

Answer» Correct Answer - d
When bead is at `A` the force it applies on the wire is radially inwards When it reaches at `B` it strikes the surface and pushed outwards Thus force is radially outwards .
387.

A rocket of initial mass `6000kg` ejects mass at a constant rate of `16kg//s` with constant relative speed of `11m//s` What is the acceleration of the rocket one mnute after blast ?A. `25"ms"^(-2)`B. `50"ms"^(-2)`C. `10"ms"^(-2)`D. `35"ms"^(-2)`

Answer» Correct Answer - D
Acceleration of the rocket at any instant t is `a=(v_(r)(dm)/(dt))/(m-t(dm)/(dt))`
Here, M=6000kg, `(dm)/(dt)=16kgs^(-1)`
`v_(r)=11kms^(-1)=11000ms^(-1), t=min =60s`
`therefore a=[(11000xx16)/(6000-60xx16)]ms^(-)approx 35ms^(-2)`
388.

A helicopter of mass 1000 kg rises with a vertical acceleration of `15 ms^(-2)` The crew and the passengers weight 300 kg Give the magnitude and direction of (a) force on the floor by the the crew and passengers (b) action of the rotor of the helicopter on surrounding air (c) force on the helicopter due to surrounding air .

Answer» Here , mass of helicopter , ` m_(1) = 1000 kg `
Mass of the crew & passengers , ` m_(2) = 300 kg `
upward acceleration , ` a = 15 ms^(-2) and g = 10^(-2) `
(a) Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers ` = m_(2) (g+ a ) = 300 (10 + 15 ) = 750 N `
(b) Action of rotor of helicopter on surrounding air is obviously vertically downwards , because helicopter rises on account of reaction to this force of action
` F = (m_(1) + m_(2)) (g + a ) = (1000 + 300) (10 + 15 ) = 1300 xx 25 = 32500 N `
(c) Force on the helicopter due to surrounding air is the reaction As action and reaction are equal and opposite therefore , Force of reaction , F = 32500 N , verically upwards .
389.

A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative] Lowest point Highest point `T_(1)` and `V_(1)` denote the tension and speed at the lowest point `T_(2)` and `V_(2)` denote the corresponding values at the highest points.

Answer» Correct Answer - A
At the lowest point, mg acts downward as `T_(1)` upwards so that net force=`mg+T_(2)`. Hence option (a) is correct.
390.

A bucket containing water is tied to one end of a rope of length 2.5 m and rotated about the other end in a vertical circle so that water does not spill even when bucet is upside down . What is the maximum velocity of the bucket at which this happens ? How many rotations per minute is it making `g = 10 m//s^(2) ` .

Answer» Here, ` = 2.5 m`.Water in the bucket will not spill when weight of water is balanced by
centrifugal force , `mg = (m upsilon^(2))/(r)`,
`upsilon = sqrt(rg) = sqrt(2.5 xx 10 ) = 5 m//s `
This is the velocity at the highest point
Angular speed , `omega = (upsilon)/(r) = (5)/(2.5) = 2 rad //s `
Frequency of rotation `n = (omega)/(2pi) = (2)/(2 pi) rps = (60)/(pi) "rpm"`
391.

A stone is tied to one end of a string and rotated in a vertical circle . What is the difference in tensions of the string at lowest and highest points of the vertical circle ?

Answer» `T_(L) - T_(H) = 6 mg = 6` times the weight of the stone .
392.

A body of mass `2kg` travels according to the law `x (t) = pt + qt^(2) + rt^(3)` where `p = 3ms^(-1),q = 4 ms^(-2)` and `r = 5ms^(-3)` .A. `136N`B. `134N`C. `158N`D. `68N`

Answer» Correct Answer - a
Here, `m = 2kg`
`x (t) = pt + qt^(2) + rt^(3)`
`upsilon = (dx)/(dt) = p + 2 qt + 3rt^(2)`
`a = (d upsilon)/(dt) = 0 + 2q + 6rt`
As `t =2 sec, a = 2q + 12 r = 2xx4 + 12 xx5 =68m//s^(2)`
`F =ma = 2 xx 68 N = 136 N` .
393.

A body of mass 2kg travels according to the law x(t) = `pt + qt^(2) + rt^(3)` where , q = `4 ms^(-2)` , p = `3 ms^(-1)` and `r = 5 ms^(-3)`. The force acting on the body at t = 2s isA. 136 NB. 134 NC. 158 ND. 68 N

Answer» Correct Answer - A
(a) Given , mass =2 kg
`x(t)=pt+qt^(2)+rt^(3)`
`v=(dx)/(dt)=p+2qt+3rt^(2)`
`a=(dv)/(dt)=0+2q+6rt`
At` t=2s,a=2q+6xx2xxr=2q+12r`
=`2xx4+12xx5=8+60=68 ms^(-1)`
Force ,` F=ma=2xx68=136N`
394.

A body with mass 5 kg is acted upon by a force `vec(F) = (- 3 hat (i) + 4 hat (j)) N`. If its initial velocity at t =0 is `vec(v) = 6 hat(i) - 12 hat (j) ms^(-1)`, the time at which it will just have a velocity along the y-axis is :A. neverB. 10sC. 2sD. 15s

Answer» Correct Answer - B
(b) Given mass, m=5 kg
Acting force,`F=(-3 hati+4 hatj)ms^(-1)`
Initial velocity at t=0,u=`(6 hati+12 hatj)ms^(-1)`
Retardation, `capa=(F)/(m)=(-(3hati)/(5)+(4hatj)/(5))ms^(-2)`
As final velocity is along Y-axis only, its x-component must be zero.
From v=u+at, for X-component only, `0=6hati-(3hati)/(5)t`
`t=(5xx6)/(3)=10s`
395.

Acceleration of a body can be……. Only if…….

Answer» non - zero unbalanced exernal force acting on the body is non-zero .
396.

The inability of a body….. Its state….. Or……is called…….of the body.

Answer» to change , of rest , state of motion with constant velocity , inertia .
397.

Three blocks with masses `m` , `2m` and `3m` are connected by strings, as shown in the figure. After an upward force `F` is applied on block `m` , the masses move upward at costant speed `v` . What is the net force on the block of mass `2m` ? (`g` is the acceleration due to gravity). ltBrgt A. zeroB. 2 mgC. 3 mgD. 6 mg

Answer» Correct Answer - A
(a) Since, all the blocks are moving with constant velocity, then the force on the all blocks will be zero.
398.

An initially stationary device lying on a frictionless floor explodes into two pieces and slides across the floor. One piece is moving in positive x-direction then other piece is moving inA. positive y- directionB. negative y-directionC. negative x-directionD. at angle from x-direction

Answer» Correct Answer - C
( c) From law of conservation of momentum,
`p_(i)=p_(f)`
and initial momentum, `p_(i)=mu=m(0)=0`
`therefore p_(f)` should also be zero.
Hence, other piece will move in negative x-direction.
399.

A bullet mass 10gm is fired from a gun of mass 1 kg . If the recoil velocity is 5 `m//s`, the velocity of the muzzle isA. 30km/minB. 60 km/minC. 30m/sD. 500m/s

Answer» Correct Answer - D
(d) Conservation of linear momentum gives
`m_(1)v_(1)=-m_(2)v_(2) implies v_(1)=(-m_(2)v_(2))/(m_(1))`
Given, `m_(1)=10 kg= ((10)/(1000)) kg`
`m_(2)=1 kg " and " v_(2)=-5 m//s`
`therefore` Velocity of muzzle,
`v_(1)=(+1xx5)/(10//1000)=500 m//s`
400.

A shell of mass 200g is fired by a gun of mass 100kg. If the muzzle speed of the shell is `80ms^(-1)`, then the recoil speed of the gun is

Answer» m = 0.02 kg , M = 100 kg , v = `80 ms^(-1)` , V = ?
Negative sign indicates that the gun moves in a direction opposite to the direction of motion of the bullet.