question_answer

An alternating voltage \[\text{v}(t)=220\sin 100\pi t\] volt is applied to a purely resistance load of \[50\Omega .\] The time taken for the current to rise from half of the peak value to the peak value is :                         [JEE Main 8-4-2019 Morning]

A) 2.2 ms             

B) 5 ms

C) 3.3 ms

D) 7.2 ms

Correct Answer: C

Solution :

Suppose 'm' gram of water evaporates then, heat required \[\Delta {{Q}_{req}}=m{{L}_{\text{v}}}\] Mass that converts into ice = (150 - m) So, heat released in this process \[\Delta {{Q}_{rel}}=(150-m){{L}_{f}}\] Now, \[\Delta {{Q}_{rel}}=\Delta {{Q}_{req}}\] \[(150-m){{L}_{f}}=m{{L}_{V}}\] \[m({{L}_{f}}+{{L}_{\text{v}}})=150{{L}_{f}}\] \[m=\frac{150{{L}_{f}}}{{{L}_{f}}+{{L}_{\text{v}}}}\] \[m=20g\] \[V\left( t \right)=220\text{ }sin(100\pi t)\text{ }volt\] time taken, \[t=\frac{\theta }{\omega }=\frac{\frac{\pi }{3}}{100\pi }=\frac{1}{300}\sec \]\[=3.3ms\]