Time taken by a $$836\,W$$ heater to heat one litre of water from $${10^ \circ }C$$ to $${40^ \circ }C$$ is
A.
$$150\,s$$
B.
$$100\,s$$
C.
$$50\,s$$
D.
$$200\,s$$
Answer :
$$150\,s$$
Solution :
$$\eqalign{
& \Delta Q = mC \times \Delta T \cr
& = 1 \times 4180 \times \left( {40 - 10} \right) = 4180 \times 30 \cr} $$
($$\therefore $$ $$\Delta Q =$$ heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \circ }C$$ to $${40^ \circ }C$$ )
$${\text{also}}\,\Delta Q = 836 \times t \Rightarrow t = \frac{{4180 \times 30}}{{836}} = 150s$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.