JEE Advance - Physics (2016 - Paper 1 Offline - No. 4)
The position vector $$\overrightarrow r $$ of a particle of mass m is given by the following equation
$$$\overrightarrow r \left( t \right) = \alpha {t^3}\widehat i + \beta {t^2}\widehat j,$$$where $$\alpha = {{10} \over 3}m{s^{ - 3}}$$, $$\beta = 5\,m{s^{ - 2}}$$ and m = 0.1 kg. At t = 1 s, which of the following
statement(s) is(are) true about the particle?
The velocity $$\overrightarrow v $$ is given by $$\overrightarrow v = \left( {10\widehat i + 10\widehat j} \right)$$ ms-1
The angular momentum $$\overrightarrow L $$
with respect to the origin is given by $$\overrightarrow L = - \left( {{5 \over 3}} \right)\widehat k\,N\,m\,s$$
The force $$\overrightarrow F $$
is given by $$\overrightarrow F = \left( {\widehat i + 2\widehat j} \right)N$$
The torque $$\overrightarrow \tau $$ with respect to the origin is given by $$\overrightarrow \tau = - \left( {{{20} \over 3}} \right)\widehat k\,N\,m$$
Explanation
$$r = \alpha {t^3}\widehat i + \beta {t^2}\widehat j$$
$$v = {{dr} \over {dt}} = 3\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$$
$$a = {{{d^2}r} \over {d{t^2}}} = 6\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$$
At t = 1 s,
(a) $$v = 3 \times {{10} \over 3} \times 1\widehat i + 2 \times 5 \times 1\widehat j$$
$$ = (10\widehat i + 10\widehat j)$$ m/s
(b) $$\widehat L = \widehat r \times \widehat p$$
$$ = \left( {{{10} \over 3} \times 1\widehat i + 5 \times 1\widehat j} \right) \times 0.1(10\widehat i + 10\widehat j)$$
$$ = \left( { - {5 \over 3}\widehat k} \right)$$ N-ms
(c) $$F = ma$$
$$ = m \times \left( {6 \times {{10} \over 3} \times 1\widehat i + 2 \times 5\widehat j} \right) = (2\widehat i + \widehat j)N$$
(d) $$\tau = r \times F = \left( {{{10} \over 3}\widehat i + 5\widehat j} \right) \times (2\widehat i + \widehat j)$$
$$ = + {{10} \over 3}\widehat k + 10( - \widehat k) = \left( { - {{20} \over 3}\widehat k} \right)$$ N-m
$$v = {{dr} \over {dt}} = 3\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$$
$$a = {{{d^2}r} \over {d{t^2}}} = 6\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j$$
At t = 1 s,
(a) $$v = 3 \times {{10} \over 3} \times 1\widehat i + 2 \times 5 \times 1\widehat j$$
$$ = (10\widehat i + 10\widehat j)$$ m/s
(b) $$\widehat L = \widehat r \times \widehat p$$
$$ = \left( {{{10} \over 3} \times 1\widehat i + 5 \times 1\widehat j} \right) \times 0.1(10\widehat i + 10\widehat j)$$
$$ = \left( { - {5 \over 3}\widehat k} \right)$$ N-ms
(c) $$F = ma$$
$$ = m \times \left( {6 \times {{10} \over 3} \times 1\widehat i + 2 \times 5\widehat j} \right) = (2\widehat i + \widehat j)N$$
(d) $$\tau = r \times F = \left( {{{10} \over 3}\widehat i + 5\widehat j} \right) \times (2\widehat i + \widehat j)$$
$$ = + {{10} \over 3}\widehat k + 10( - \widehat k) = \left( { - {{20} \over 3}\widehat k} \right)$$ N-m
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