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JEE Mains 2020 Model Paper 1 New Pattern

PHYSICS

(SINGLE CORRECT ANSWER TYPE)

This section contains 20 multiple choice questions. Each question has 4 options (1), (2), (3) and (4) for its answer, out of which ONLY ONE option can be correct.

Question : 26 of 75

Marks: +1, -0

The lower end of a 4 m long uniform rod AB is pulled with constant speed υ = 4 m/ s . The speed of centre of mass of the rod at θ = 60° will be

$\frac{4}{\sqrt{3}}$ m/s
$2\sqrt{3}$ m/s
4 m/s
$4\sqrt{3}$ m/s

Validate

Solution:

OB = x (say) = AB cos θ

Or x = 4 cos θ

∴

\frac{dx}{dt}

(-4\sin\theta)\frac{d\theta}{dt}

But

\frac{dx}{dt}

= 4 m/s

∴

\left(-\frac{d\theta}{dt}\right)_{\text{at } 60^{\circ}}

= cosec 60° =

\frac{2}{\sqrt{3}}

rad/s

Further,

\overset{\rightarrow}{r_c}

(2 \cos \theta)\hat{i} + (2 \sin \theta)\hat{j}

\overset{\rightarrow}{r_c}

\frac{\overset{\rightarrow}{dr}}{dt}

\left(-2 \sin\theta \cdot \frac{d\theta}{dt}\right) \hat{\mathbf{i}} + \left(2 \cos\theta \cdot \frac{d\theta}{dt}\right) \hat{\mathbf{j}}

\overset{\rightarrow}{v_c}

\left(2 \times \frac{\sqrt{3}}{2} \times \frac{2}{\sqrt{3}}\right) \hat{i} + \left[ (2)\left(\frac{1}{2}\right)\left(-\frac{2}{\sqrt{3}}\right) \right] \hat{j}

2 i^{\hat{\mathbf{something}}} - \frac{2}{\sqrt{3}} j^{\hat{\mathbf{something}}}

∴

v_c

\sqrt{4+\frac{4}{3}}

\frac{4}{\sqrt{3}}

m/s

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