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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Laws of Motion

BITSAT 2021 · Physics · Laws of Motion

A ball of mass 0.2 ~kg is thrown vertically upwards by applying a...

A ball of mass 0.2 kg0.2 \mathrm{~kg} is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m0.2 \mathrm{~m} while applying the force and the ball goes upto 2 m2 \mathrm{~m} height further, find the magnitude of the force. (Consider \left.g=10 \mathrm{~m} / \mathrms^{2}\right)$.

Options

  1. A4N4 \mathrm{ N}
  2. B16N16 \mathrm{ N}
  3. C20N20 \mathrm{ N}
  4. D22N22 \mathrm{ N}

Correct answer

D. 22N22 \mathrm{ N}

Step-by-step solution

Let the velocity of the ball just when it leaves the hand is uu then applying, $\begin{array}{l} v^{2}-u^{2}=2 a \text{ for upward journey } \\ \Rightarrow-u^{2}=2(-10) \times 2 \\ \Rightarrow u^{2}=40 \end{array} Again applying v2−u2=2asv^{2}-u^{2}=2 a s for the upward journey of the ball, when the ball is in the hands of the thrower, v2−u2=2as⇒40−0=2(a)0.2⇒a=100 m/\mathrms2∴F=ma=0.2×100=20 N⇒N−mg=20⇒N=20+2=22 N\begin{array}{l}v^{2}-u^{2}=2 a s \\ \Rightarrow 40-0=2(a) 0.2 \\ \Rightarrow a=100 \mathrm{~m}/ \mathrms^{2} \\ \therefore \mathrm{F}=m a=0.2 \times100=20 \mathrm{~N}\\ \Rightarrow \mathrm{N}-m g=20 \\ \Rightarrow \mathrm{N}=20+2=22 \mathrm{~N} \end{array}
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