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When a body rolls down an inclined plane (θ) without slipping the velocity on reaching the ground is,

\(v= \sqrt \frac{2gh}{1+\frac{k^2}{r^2}} \)

where h is the vertical height of inclined plane and K is the radius of gyration of the rolling body.

Work done in rotational motion,

W = ∑ F_ir_i × ∆θ,

where ∑ F_ir_i is the algebraic sum of moment of force and ∆θ is the angle through which a body is rotated,

∴ W = Total torque × Angular displacement

We know that τ = rFsin θ.

If θ = 0 or 180° , or r = 0, then τ = 0. r = 0 means the applied force passes through the axis of rotation.

No, a torque is required only for producing angular acceleration.

It is false : Since the body is rotating, its instantaneous acceleration is not zeor.

It is true : A rolling body can be magined to be rotating about an axis passing through the point of contact of the of the body with the ground and hence its instantaneous speed is zero.

Let a physical quantity be given by

A = n₁u₁ = n₂u₂

where n₁ is numerical value with u₁ as unit in one system and similarly n₂ is numerical value with u₂ as unit in another system.

If dimensional formula of A is [M^aL^bT^c ]

then, n₁ [M₁^aL₁^bT₁^c ] = n₂[M₂^aL₂^bT₂^c ]

i.e.,   \(n_2 = n_1\frac{[M_1^aL_1^bT_1^c]}{[M_1^aL_1^bT_1^c]}\)

= \(n_1[\frac{M_1}{M_2}]^a\)\([\frac{L_1}{L_2}]^b\)\([\frac{T_1}{T_2}]^c\)

We know that

I₁ω₁ = I₂ω₂

or \(mk^2_1ω_1=mk^2_2w_2\)

or \((\frac{k_1}{k_2})^2=\frac{ω_2}{ω_1}\)

The turning effect of a force is called Torque. The angular momentum of the body is defined as the moment of momentum about the axis of rotation. The relation between Torque & angular momentum is L = \(\frac {τw}{a}.\)

(a) iii, 

(b) iv 

(c) ii 

(d) i.

The angular momentum of the particle is defined as the moment of momentum about the axis of rotation.

∴ Angular momentum = linear momentum x perpendicular distance from the axis.

If ω is the angular velocity of the particle, then the linear velocity v = r ω.

∴ Angular momentum of each particle = (mv)r = (mrω) r = mr²ω.

∴ Total angular momnetum of the body is L = (Σ mr²ω) = (Σ mr²)

ω = lω where l = Σ mr².

(a) ii, 

(b) iii, 

(c) i, 

(d) iv

Moment of inertia of a solid sphere about the diameter I = \(\frac{2}{5}\)MR². Since M & R is the same. I depend on R. Since R is less for ironwood has a greater moment of inertia.

The hollow sphere shall have greater M.I., as its entire mass is concentrated at the boundary of the sphere at maximum distance from the axis.

Same work is done in both cases.

M.I. of sphere about its axis = \(\frac{2}{5}\)MR²

M.I. of hollow cylinder about its axis = MR²

Since torque to both bodies are the same

τ  = 1 α ⇒ α = \(\frac {τ}{1}\) 

Hence the object with lesser M.I. will have a higher angular acceleration and hence higher angular speeds at any moment. Thus the sphere will have higher angular speeds after a given time.

M.I of the cylinder = I₁ = MR²

and M.I. of the sphere = I₂ = 2/5 MR²

Angular acceleration of the cylinder,

α₁ = τ/I₁ = {τ}/{MR₂}

Angular acceleration of the sphere,

α₂ = τ/I₂ = {τ}/{2/5 MR₂} = {2.5 τ}/{MR₂} = 2.5 α₁

It is clear that the sphere will acquire a greater angular acceleration.

Positive slope indicates anticlockwise rotation which is traditionally taken as positive

M.I. is not affected by speed of rotation of the body.

The sphere of metal with smaller density shall be bigger in size and hence it will have large M.I.

I = Σm_ir_i². All the mass in a cylinder lies at distance R from theaxis of symmetry but most of the mass of a solid sphere lies at a smaller distance than R.

The minimum MI will be about an axis passing through the centre of the cube and connecting the opposite corners.

I = \(\frac{1}{2}\)MR² , where M is the mass and R is radius of the hollow sphere.

Correct answer is (B) Uniform motion in a horizontal circle

Using theorem of parallel axis

\(I' = I'_{CM}+Mα^2\)

If a = l/2, then

\(I' = \frac{Ml^2}{12}+M(\frac{l}{2})^2\)

\(=\frac{Ml^2}{3}\)

Using theorem of perpendicular axes

I_z = I_x + I_y

As I_x and I_y are along the two diameters of disc so using symmetery,

I_x = I_y

So I_z = 2I_x But I_z = \(\frac{MR^2}{2}\)

So I_x = \(\frac{I_z}{2}\) = \(\frac{MR^2}{4}\)

The tangent is parallel to its radius. The distance between two axes is R. Using theorem of parallel axes.

I_tangent = I_diameter + MR²

= \(\frac{MR^2}{2}+ MR^2 =\frac{3}{2}MR^2\)

1. Moment of inertia of a thin rod of length L and mass M about an axis passing through its centre and perpendicular to its length is, I = \(\frac {ML^2}{2}\)

2. Moment of inertia of a circular ring of radius R and mass M about an axis passing through the diameter is,l = \(\frac {ML^2}{2}\)

Moment of inertia of a circular ring of radius R and mass M about an axis passing through the centre and perpendicular to its plane is, I = MR².

3. Moment of inertia of a cylinder of radius R and length L and mass M about an axis passing through the axis is I =\(\frac {ML^2}{2}\)

4. Moment of inertia of a solid sphere of radius R and mass M passing through the diameter is I = \(\frac {2}{5}\)MR².

(C) less than that of a simple pendulum of same length l.

a. Initial angular velocity ω₁ = 2πn₁

Here n₁ = \(\frac {1200}{60}\) = 20 rps

ω₁ = 2πn₁

= 2 × π × 20 = 40π rads^-1

Final angular velocity ω₂ = 2πn₂

Here n₂ = \(\frac {600}{60}\) = 10 rps

∴ ω₂ = 2 × π × n₂

= 2 × π × 10 = 20π rads^-1

Angular acceleration

a = \(\frac {ω_2-ω_1}{t}\) = \(\frac {20π - 40π}{5}\) \(= \frac{-20π}{5}\)

= -4π

i.e. a = 12.57 rad s^-2. Therefore angular retardation is 12.57 rad s² .

Correct Option is (B) \(45^\circ\)

Range of projectile ,

\(R = \frac{u^2 \ sin2 \ \theta}{g}\)

\(R_{max} = \frac{u^2 \ sin2 \ \theta}{g}\)

\(\because \theta = 45^\circ\)

\(R_{max} = \frac{u^2 \ sin 90^\circ}{g}\)

\(R_{max} = \frac{u^2}{g}\)

Correct option is: (B) 45°

Correct Option is (C) Energy

We know that, average kinetic energy E = \(\frac{3}{2} K_BT\)

It means energy does not depend on the mass of molecule

Correct option is (C) energy

The basic assumptions of the kinetic theory of an ideal gas :

1. A gas of a pure material consists of an extremely large number of identical molecules. 

2. A gas molecule behaves as an ideal particle, i.e., it has mass but its structure and size can be ignored as compared with the intermolecular separation in a dilute gas and the dimensions of the container. 

3. The molecules are in constant random motion with various velocities and obey Newton’s laws of motion. 

4. Intermolecular forces can be ignored on the average so that the only forces between the molecules and the walls of the container are contact forces during collisions. It follows that between successive collisions, a gas molecule travels in a straight line with constant speed.

5. The collisions are perfectly elastic conserving total momentum and kinetic energy, and the duration of a collision is very small compared to the time interval between successive collisions.

Notes :

(1) The walls of the container holding the gas are assumed to be rigid and infinitely massive so that they do not move. 

(2) Assumption (2) allows us to ignore intermolecular collisions because if they are truly point like (i.e., of negligible extent) they cannot make contact with each other. Therefore, the only collisions we consider are those with the walls of the container. If these collisions are perfectly elastic [by assumption (5)], they only change the direction of the velocity of a gas molecule and a gas molecule possesses only kinetic energy by assumption

(4). The kinetic theory of gases was developed by Daniel Bernoulli (1700 – 82), Swiss mathematician, Rudolf Clausius (1822 – 88), German theoretical physicist, James Clerk Maxwell (1831-79), British physicist, Ludwig Eduard Boltzmann, Josiah Willard Gibbs (1839-1903), US physical chemist.]

Boyle’s law : At a constant temperature, the pressure exerted by a fixed mass of gas is inversely proportional to its volume. If P and V denote the pressure and volume of a fixed mass of gas, then, PV = constant at a constant temperature, for a fixed mass of gas. According to the kinetic theory of gases, the pressure exerted by the gas is

P = \(\frac13\)\(\frac{Nmv^2_{rms}}V\)

where N is the number of molecules of the gas, m is the mass of a single molecule, v_rms is the rms speed of the molecules and V is the volume occupied by the gas.

∴ PV = (\(\frac12\)mv² _rms) × \(\frac23\)N

= (KE of a gas molecule) \(\frac23\)N … (1)

For a fixed mass of gas, N is constant. Further, intermolecular forces are ignored so that the corresponding potential energy of the gas molecules may be assumed to be zero. Therefore, \(\frac12\)mv² _rms is the total energy of a gas molecule and N(\(\frac12\)mv²_rms) is the total energy of the gas molecules, which is proportional to the absolute temperature of the gas. Then, the right-hand side of EQ. (1) will be constant if its temperature is constant. Hence, it follows that PV = constant for a fixed mass of gas at constant temperature, which is Boyle’s law.

Correct answer is (c) falls

The system consists of three identical ball-bearings marked as 1,2 and 3. Let m be the mass of each ball-bearing.

Total kinetic energy of the system before collision

= 1/2 mv² + 0 + 0 = 1/2 mv²

Case (i) K.E. of the system after collision

= 0 + 1/2 (2m)(v/2)² = 1/4 mv²

Case (ii) K.E. of the system after collision

= 0 + 1/2 mv² = 1/2 mv²

Case (iii) K.E. of the system after collision

= 1/2 (3m)(v/3)² = 1/6 mv²

As in an elastic collision, the kinetic energy of the system remains unchanged case (ii) is the only possible result of the collision.

It states that equal volumes of all gases under similar conditions of temperature and pressure, contain equal number of molecules.

Answer is (B) For negligible duration.

Charges whose sizes are very small compared to the distance between them are called point charges. 

Net electric field in region

Direction is away from the two sheets i.e., towards right side.

For region II,

\(E_{II} = \frac{1}{2 \varepsilon_0}(\sigma_1 - \sigma_2)\)

towards right side/from sheet A to sheet B.

Foe region III,

\(E_{III} = \frac{1}{2 \varepsilon_0}(\sigma_1 + \sigma_2)\)

towards right side/away from the two sheets.

C) Both ‘A’ and ‘B’

Correct option is B) basic

A) A and R are true R does not support A

Correct option is A) acidic

Correct option is B) hydrogen

Correct option is A) Gold

Correct option is C) Metals

B) Non-metals

Correct option is D) Gold