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Byju's Answer

Standard IX

Physics

Law of Conservation of Energy

A piece of le...

Question

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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Similar questions

Q. A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. The temperature of the lead piece immediately after collision increases by (Sp.heat of lead is 30.6cal/kg/

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A metal ball of mass 200 g falls from a height of 5 m and hits a slab kept on the ground. If all the kinetic energy of the ball was imparted to the slab as heat energy, what is the rise in temperature of the slab? (Mass of slab : 5 kg, Specific heat capacity of slab = $2 J / k g^{\circ} C, g = 10 m / s^{2}$ )

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Q. A lead ball at

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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