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A block of mass $m(=0.1 \mathrm{kg})$ is hanging over a frictionless light fixed pulley by an inextensible string of negligible mass. The other end of the string is pulled by a constant force $F$ in the vertically downward direction. The linear momentum of the block increases by $2 \mathrm{kgms}^{-1}$ in $1 \mathrm{s}$ after the block starts from rest. Then, (given $g=10 \mathrm{ms}^{-2}$ ) 
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The correct answers are:
The tension in the string is $F$, The tension in the string is $3 \mathrm{N}$, The work done against the force of gravity is $10 \mathrm{J}$
The free body diagram 
$\Rightarrow \quad F=2+m g=3 N$
also, $a=\frac{\text { unbalanced force }}{\text { mass }}=\frac{2}{0.1}=20 \mathrm{m} / \mathrm{s}^{2}$
$S=\frac{1}{2} a t^{2}=\frac{1}{2} \times 20 \times 1=10 \mathrm{m}$
Honce, work done by tension
$$
=F \times 10=3 \times 10=30 \mathrm{J}
$$
So, the work done against gravity
$$
=m g \times S=1 \times 10=10 \mathrm{J}
$$
$\Rightarrow \quad F=2+m g=3 N$
also, $a=\frac{\text { unbalanced force }}{\text { mass }}=\frac{2}{0.1}=20 \mathrm{m} / \mathrm{s}^{2}$
$S=\frac{1}{2} a t^{2}=\frac{1}{2} \times 20 \times 1=10 \mathrm{m}$
Honce, work done by tension
$$
=F \times 10=3 \times 10=30 \mathrm{J}
$$
So, the work done against gravity
$$
=m g \times S=1 \times 10=10 \mathrm{J}
$$
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