A piece of lead falls from height of $100 m$ on a fixed non-conducting slab which brings it to rest. if the specific heat of lead is $30.6 c a l / k g^{\circ} C$ the increase in temperature of the slab immediately after collision is

{6.72}^{\circ} C

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{7.62}^{\circ} C

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{5.62}^{\circ} C

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{8.72}^{\circ} C

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Solution

The correct option is B ${7.62}^{\circ} C$
Given:
Height, h =100m
Specific heat of lead, s =30.6 cal/Kg°C
Solution:
Potential energy of lead = Thermal energy of slab
$\Rightarrow m g h = m s Δ T$
$\Rightarrow Δ T = \frac{g h}{s}$
$\Rightarrow Δ T = \frac{9.8 \times 100}{30.6 \times 4.2}$
$\Rightarrow Δ T = 7.62 ° C$
Hence, B is the correct option.

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A piece of lead falls from height of 100m on a fixed non-conducting slab which brings it to rest. if the specific heat of lead is 30.6cal/kg∘C the increase in temperature of the slab immediately after collision is

The correct option is B 7.62∘CGiven:Height, h =100mSpecific heat of lead, s =30.6 cal/Kg°CSolution:Potential energy of lead = Thermal energy of slab⇒mgh=msΔT⇒ΔT=ghs⇒ΔT=9.8×10030.6×4.2⇒ΔT=7.62°CHence, B is the correct option.

A piece of lead falls from height of $100 m$ on a fixed non-conducting slab which brings it to rest. if the specific heat of lead is $30.6 c a l / k g^{\circ} C$ the increase in temperature of the slab immediately after collision is