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An automobile moves on a road with a speed of 54 km h^-1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m². If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is

6.66 kgm²s^-2

8.58 kgm²s^-2

10.86 kgm²s^-2

2.86 kgm²s^-2

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Explanation

As velocity of an automobile,

v=54km/h

=54 x 5/18

=15m/s

Angular velocity of a vehicle, v=ω_or

=> ω_{o=v/R=15/0.45=100/3rad/s

$S o, a n g u l a r a c c e l e r a t i o n o f a n a u t o m o b i l e, α = Δ ω / t = ω_{f} - ω_{o} / t = 0 - 100 / 3 / 15 = - 100 / 45 r a d / s^{2} T h u s, a v e r a g e t o r q u e t r a n s m i t t e d b y i t s b r a k e s t o w h e e l τ = I α 3 x 100 / 45 = 6.66 k g m^{2} s^{- 2}$}

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A body of mass (4m) is lying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is

mv²

3/2mv²

2mv²

4mv²

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A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis.A massless string is wound round the cylinder with one end attached to it and other hanging freely.Tension in the string required to produce an angular acceleration of 2 rev/ s² is

25N

50N

78.5N

157N

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Explanation

The moment of inertia of a solid cylinder about its axis is (1/2)MR^2. The torque required for an angular acceleration Î± is IÎ±. Substituting the given values, we get Ï„ = (1/2) Ã— 50 Ã— (0.5)^2 Ã— 2Ï€ Ã— 2 = 157 N-m. The tension in the string provides this torque.

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The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle θ without slipping and slipping down the incline without rolling is

5:7

2:3

2:5

7:5

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An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^-1 and the second part of mass 2kg moves with 8 ms^-1 speed. If the third part flies off with 4 ms^-1 speed, then its mass is

3kg

5kg

7kg

17kg

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Explanation

(b) We have p₁+p₂+p₃=0 [∴ P=mv]
∴ 1x12i+2x8j+p₃=0

=> 12i+16j+p₃=0

=> p₂=-(12i+16j)

p₃=√(12)²+(16)²=144+256=20 kg-m/s

Now, p₃=m₃v₃

=> m₃=p₃/v₃=20/4=5kg

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A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

ring

solid sphere

hollow sphere

disc

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Explanation

(d) As V=√2gh/1+k²/r²
Given h=3v²/4g

v²=2gh/1+k²/r²=2g3v²/4g(1+k²/r²)=6gv²/4g(1+u²/v²)

1=3/2(1+k²/v²)

or 1+k²/r²=3/2 or k²/r²=3/2-1=1/2

k²=1/2r² (Equation of disc)

Hence, the object is a disc

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When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

a line perpendicular to the plane of rotation

the line making an angle of $45 °$ to the plane of rotation

the radius

the tangent to the orbit

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Explanation

We know L= $m (r \times v)$

So here, angular momentum is directed along a line perpendicular to the plane of rotation.

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Two persons of mass 55 kg and 65 kg respectively, are at the opposite ends of a boat.The length of the boat is 3.0m and weighs 100 kg.The 55 kg man walks up to the 65 kg man and sits with him.If the boat is in still water the centre of mass of the system shifts by

3.0m

2.3m

zero

0.75m

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Explanation

Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

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A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity 4 $m s^{- 1} .$ It collides with a horizontal spring of force constant 200 $N m^{- 1}$ . The maximum compression produced in the spring will be

0.5 m

0.6 m

0.7 m

0.2 m

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Explanation

Loss in KE = Gain in spring energy

$\frac{1}{2} m v^{2} [1 + \frac{K^{2}}{R^{2}}] = \frac{1}{2} k x_{m a x}^{2}$

where k is the force contant.

Given,v=4m/s,m=3kg,k=200N/m

For solid cylinder, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$∴ \frac{1}{2} \times 3 \times {(4)}^{2} [1 + \frac{1}{2}] = \frac{1}{2} \times 200 \times x_{m a x}^{2}$

The maximum compression in the spring

$x_{m a x} = 0.6 m$

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Two spheres A and B of masses $m_{1} a n d m_{2}$ respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction.The mass A moves after collision in the direction

same as that of B

opposite to that of B

$θ = \tan^{- 1} (\frac{1}{2}) t o t h e x ‐ a x i s$

$θ = \tan^{- 1} (\frac{- 1}{2}) t o t h e x ‐ a x i s$

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