A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips of the table in a very short time τ = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)

Option 1 : 0.5

**Concept:**

**Newton's second law ****for** **Rotation:**

If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration

Torque = (moment of inertia)(angular acceleration)

τ = Iα

τ = torque, around a defined axis (N∙m**)**

I = moment of inertia (kg∙m^{2})

α = angular acceleration (radians/s^{2})

**Calculation:**

In this problem, the rotational form of Newton's second law is used which states the relation between net external torque (T) and the angular acceleration (α) of a body about a fixed axis.

T = *I*α

In this law, rotational form of Newton's second law for rigid body is used.

Now, a rod of mass M and length*l*hinged at one end without friction and held horizontally is suddenly released under gravity. The angular acceleration of the rod just after the release.

\({T_{{\rm{net}}}} = \frac{{{\rm{MgL}}}}{2}\)

T_{net} = Iα

\(\frac{{{\rm{MgL}}}}{2} = {\rm{I\alpha }}\) ----(1)

The moment of inertia of the rod about an axis passing through the hinge and perpendicular to the rod is:

\({{\rm{I}}_{{\rm{end}}}} = \frac{1}{3}{\rm{M}}{{\rm{l}}^2}{\rm{\;}}\) (Derived)

Now, substituting moment of inertia in equation in (1)

\({\rm{Mg}}\frac{{\rm{l}}}{2} = \frac{{{\rm{M}}{{\rm{l}}^2}}}{3}{\rm{\alpha }}\)

\( \Rightarrow {\rm{\alpha }} = \frac{{Mgl}}{2} \times \frac{3}{{M{l^2}}}\)

\(\alpha = \frac{{3g}}{{2l}}\)

Where, g = Acceleration due to gravity = 10 m/s^{2}

l = Length = 0.31 m

On substituting the given values in angular acceleration,

\( \Rightarrow \alpha = \frac{{3 \times 10}}{{2 \times 0.31}}\)

∴ α = 48.3 rad/s^{2}

The angle is given by the formula:

θ = ωt

Where, ω = Angular velocity

and t = time

Since, we need angular acceleration and to obtain angular velocity, we need to multiply angular acceleration with time.

⇒ ω = α × τ

Where, τ = time (given, the time taken during rotation)

On substituting the values,

⇒ ω = 48.3 × 0.01

∴ ω = 0.483 rad/s

Now, the time taken to hit the ground is given by the formula:

\(t = \sqrt {\frac{{2h}}{g}} \)

Given, Height of the rod h = 5 m

On substituting the values,

\( \Rightarrow t = \sqrt {\frac{{2 \times 5}}{{10}}} \)

∴ t = 1 s

Now, the angle by which the rod would rotate is given by the formula:

θ = ωt

On substituting the values,

⇒ θ = 0.483 × 1

∴ θ = 0.483 radian ≈ 0.5 radianJEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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