A body of mass 25 kg is dragged on a rough horizontal floor for one hour with a speed of 2 kmh -1. The coefficient of friction for the surface in contact is 0.5 and half the heat produced is absorbed by the body. If specific heat of body is 0.1 cal g -1( 0 C -1) and g =9.8 ms -2, then the rise in temperature of body is

Question

A body of mass 25 kg is dragged on a rough horizontal floor for one hour with a speed of 2 kmh -1. The coefficient of friction for the surface in contact is 0.5 and half the heat produced is absorbed by the body. If specific heat of body is 0.1 cal g -1( 0 C -1) and g =9.8 ms -2, then the rise in temperature of body is (A) 39 K (B) 59.5 K (C) 84.5 K (D) 11.6 K

Tardigrade Admin · Accepted Answer

Power supplied against frictional force =(μ mg ) v Heat produced =((μ mgv ) t /2)= ms Δ θ ⇒ Δ θ=(μ g v t/2 s )=((0.5)(9.8)(2 × (5/18))(3600)/2 × 0.1 × 4.2 × 103)=11.6 K

$39 K$

$59.5 K$

$84.5 K$

$11.6 K$

Power supplied against frictional force $= (μ m g) v$
Heat produced $= \frac{( μ m gv ) t}{2} = m s Δ θ$
$\Rightarrow Δ θ = \frac{μgv t}{2 s} = \frac{( 0.5 ) ( 9.8 ) ( 2 \times \frac{5}{18} ) ( 3600 )}{2 \times 0.1 \times 4.2 \times 1 0 ^{3}} = 11.6 K$

39K

59.5K

84.5K

11.6K

Power supplied against frictional force =(μmg)v Heat produced =2(μmgv)t​=msΔθ ⇒Δθ=2sμgvt​=2×0.1×4.2×103(0.5)(9.8)(2×185​)(3600)​=11.6K

$39 K$

$59.5 K$

$84.5 K$

$11.6 K$

Power supplied against frictional force $= (μ m g) v$
Heat produced $= \frac{( μ m gv ) t}{2} = m s Δ θ$
$\Rightarrow Δ θ = \frac{μgv t}{2 s} = \frac{( 0.5 ) ( 9.8 ) ( 2 \times \frac{5}{18} ) ( 3600 )}{2 \times 0.1 \times 4.2 \times 1 0 ^{3}} = 11.6 K$