Verified Solution • 2025–26 Session moderate

QJEEAdv16Ph4cc1: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph4cc1 Solution: The position vector of a particle of mass m is given by the…

The position vector r→\overrightarrow r of a particle of mass m is given by the following equation </mo></mrow><mover><mi>r</mi><mo>→</mo></mover><mrowdata−mjx−texclass="INNER"><modata−mjx−texclass="OPEN">(</mo><mi>t</mi><modata−mjx−texclass="CLOSE">)</mo></mrow><mo>=</mo><mi>α</mi><mrowdata−mjx−texclass="ORD"><msup><mi>t</mi><mn>3</mn></msup></mrow><mrowdata−mjx−texclass="ORD"><mover><mi>i</mi><mo></mo></mover></mrow><mo>+</mo><mi>β</mi><mrowdata−mjx−texclass="ORD"><msup><mi>t</mi><mn>2</mn></msup></mrow><mrowdata−mjx−texclass="ORD"><mover><mi>j</mi><mo></mo></mover></mrow><mo>,</mo></math></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container></mjx−assistive−mml></mjx−container><span></mo></mrow><mover><mi>r</mi><mo>→</mo></mover><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>t</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mo>=</mo><mi>α</mi><mrow data-mjx-texclass="ORD"><msup><mi>t</mi><mn>3</mn></msup></mrow><mrow data-mjx-texclass="ORD"><mover><mi>i</mi><mo>^</mo></mover></mrow><mo>+</mo><mi>β</mi><mrow data-mjx-texclass="ORD"><msup><mi>t</mi><mn>2</mn></msup></mrow><mrow data-mjx-texclass="ORD"><mover><mi>j</mi><mo>^</mo></mover></mrow><mo>,</mo></math></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container></mjx-assistive-mml></mjx-container><span>where</span><span></span><span>\alpha = {{10} \over 3}m{s^{ - 3}}</span><span></span><span>,</span><span></span><span>\beta = 5\,m{s^{ - 2}}</span></span> and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?

Multiple Choice Options

  • A The velocity v→\overrightarrow v is given by v→=(10i^+10j^)\overrightarrow v = \left( {10\widehat i + 10\widehat j} \right) ms-1
  • B The angular momentum L→\overrightarrow L with respect to the origin is given by L→=−(53)k^ N m s\overrightarrow L = - \left( {{5 \over 3}} \right)\widehat k\,N\,m\,s
  • C The force F→\overrightarrow F is given by F→=(i^+2j^)N\overrightarrow F = \left( {\widehat i + 2\widehat j} \right)N
  • D The torque τ→\overrightarrow \tau with respect to the origin is given by τ→=−(203)k^ N m\overrightarrow \tau = - \left( {{{20} \over 3}} \right)\widehat k\,N\,m

Step-by-Step Verified Solution

r=αt3i^+βt2j^r = \alpha {t^3}\widehat i + \beta {t^2}\widehat j

v=drdt=3αt2i^+2βt2j^v = {{dr} \over {dt}} = 3\alpha {t^2}\widehat i + 2\beta {t^2}\widehat j

a=d2rdt2=6αt2i^+2βt2j^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=3×103×1i^+2×5×1j^v = 3 \times {{10} \over 3} \times 1\widehat i + 2 \times 5 \times 1\widehat j

=(10i^+10j^)= (10\widehat i + 10\widehat j) m/s

(b) L^=r^×p^\widehat L = \widehat r \times \widehat p

=(103×1i^+5×1j^)×0.1(10i^+10j^)= \left( {{{10} \over 3} \times 1\widehat i + 5 \times 1\widehat j} \right) \times 0.1(10\widehat i + 10\widehat j)

=(−53k^)= \left( { - {5 \over 3}\widehat k} \right) N-ms

(c) F=maF = ma

=m×(6×103×1i^+2×5j^)=(2i^+j^)N= 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) τ=r×F=(103i^+5j^)×(2i^+j^)\tau = r \times F = \left( {{{10} \over 3}\widehat i + 5\widehat j} \right) \times (2\widehat i + \widehat j)

=+103k^+10(−k^)=(−203k^)= + {{10} \over 3}\widehat k + 10( - \widehat k) = \left( { - {{20} \over 3}\widehat k} \right) N-m

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