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Standard XII

Physics

Longitudinal Stress

A uniform rod...

Question

A uniform rod of mass $6 k g$ is lying on a smooth horizontal surface. Its two ends are pulled by strings as shown in figure. Force exerted by $40 c m$ part of the rod on $10 c m$ part of the rod is

30 N

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16 N

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24 N

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22 N

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Solution

The correct option is B $16 N$

Applying $F_{n e t} = m a$ for the entire rod
$40 - 10 = 6 a$
$\Rightarrow a = 5$
FBD of part A

( $T$ is the force on $10 c m$ part exerted by the $40 c m$ part)

$T - 10 = m_{A} a (m_{A} = \frac{6}{50} \times 10 k g)$
$T = \frac{6}{5} \times 5 + 10 = 16 N$

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A uniform rod of mass 6 kg is lying on a smooth horizontal surface. Its two ends are pulled by strings as shown in figure. Force exerted by 40 cm part of the rod on 10 cm part of the rod is

The correct option is B 16 N Applying Fnet=ma for the entire rod 40−10=6a ⇒a=5 FBD of part A (T is the force on 10 cm part exerted by the 40 cm part) T−10=mAa(mA=650×10 kg) T=65×5+10=16 N

A uniform rod of mass $6 k g$ is lying on a smooth horizontal surface. Its two ends are pulled by strings as shown in figure. Force exerted by $40 c m$ part of the rod on $10 c m$ part of the rod is

The correct option is B $16 N$

Applying $F_{n e t} = m a$ for the entire rod
$40 - 10 = 6 a$
$\Rightarrow a = 5$
FBD of part A

( $T$ is the force on $10 c m$ part exerted by the $40 c m$ part)

$T - 10 = m_{A} a (m_{A} = \frac{6}{50} \times 10 k g)$
$T = \frac{6}{5} \times 5 + 10 = 16 N$