01. Potential and Kinetic Energy
Work, Energy and Power

148930 If the kinetic energy of a body becomes four times then its momentum will be

1 $p_{\text {new }}=3 p_{\text {initial }}$
2 $\mathrm{p}_{\text {new }}=4 \mathrm{p}_{\text {initial }}$
3 $\mathrm{p}_{\text {new }}=2 \mathrm{p}_{\text {initial }}$
4 $p_{\text {new }}=p_{\text {initial }}$
Work, Energy and Power

148931 In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. the expression for kinetic energy of block when it reaches point ' $B$ ' is

1 $\frac{1}{2} \mathrm{mgy}_{0}^{2}$
2 $\frac{1}{2} \mathrm{mgy}^{2}$
3 $\operatorname{mg}\left(\mathrm{y}-\mathrm{y}_{0}\right)$
4 $\mathrm{mgy}_{0}$
Work, Energy and Power

148932 Water falls from a $40 \mathrm{~m}$ high dam at the rate of $9 \times 10^{4} \mathrm{~kg}$ per hour. Fifty percent of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of $100 \mathrm{~W}$ lamps, that can be lit, is:
(Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

1 25
2 50
3 100
4 18
Work, Energy and Power

148933 Position of a $3 \mathrm{~kg}$ mass moving along the $x$-axis is given by $x=0.3 \cos (\omega t) \mathrm{m}$. If $K(t)$ denotes the kinetic energy at time $t$,
Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

1 $\frac{1}{3}$
2 $\frac{1}{2}$
3 $\frac{\sqrt{3}}{2}$
4 $\sqrt{3}$
Work, Energy and Power

148930 If the kinetic energy of a body becomes four times then its momentum will be

1 $p_{\text {new }}=3 p_{\text {initial }}$
2 $\mathrm{p}_{\text {new }}=4 \mathrm{p}_{\text {initial }}$
3 $\mathrm{p}_{\text {new }}=2 \mathrm{p}_{\text {initial }}$
4 $p_{\text {new }}=p_{\text {initial }}$
Work, Energy and Power

148931 In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. the expression for kinetic energy of block when it reaches point ' $B$ ' is

1 $\frac{1}{2} \mathrm{mgy}_{0}^{2}$
2 $\frac{1}{2} \mathrm{mgy}^{2}$
3 $\operatorname{mg}\left(\mathrm{y}-\mathrm{y}_{0}\right)$
4 $\mathrm{mgy}_{0}$
Work, Energy and Power

148932 Water falls from a $40 \mathrm{~m}$ high dam at the rate of $9 \times 10^{4} \mathrm{~kg}$ per hour. Fifty percent of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of $100 \mathrm{~W}$ lamps, that can be lit, is:
(Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

1 25
2 50
3 100
4 18
Work, Energy and Power

148933 Position of a $3 \mathrm{~kg}$ mass moving along the $x$-axis is given by $x=0.3 \cos (\omega t) \mathrm{m}$. If $K(t)$ denotes the kinetic energy at time $t$,
Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

1 $\frac{1}{3}$
2 $\frac{1}{2}$
3 $\frac{\sqrt{3}}{2}$
4 $\sqrt{3}$
Work, Energy and Power

148930 If the kinetic energy of a body becomes four times then its momentum will be

1 $p_{\text {new }}=3 p_{\text {initial }}$
2 $\mathrm{p}_{\text {new }}=4 \mathrm{p}_{\text {initial }}$
3 $\mathrm{p}_{\text {new }}=2 \mathrm{p}_{\text {initial }}$
4 $p_{\text {new }}=p_{\text {initial }}$
Work, Energy and Power

148931 In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. the expression for kinetic energy of block when it reaches point ' $B$ ' is

1 $\frac{1}{2} \mathrm{mgy}_{0}^{2}$
2 $\frac{1}{2} \mathrm{mgy}^{2}$
3 $\operatorname{mg}\left(\mathrm{y}-\mathrm{y}_{0}\right)$
4 $\mathrm{mgy}_{0}$
Work, Energy and Power

148932 Water falls from a $40 \mathrm{~m}$ high dam at the rate of $9 \times 10^{4} \mathrm{~kg}$ per hour. Fifty percent of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of $100 \mathrm{~W}$ lamps, that can be lit, is:
(Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

1 25
2 50
3 100
4 18
Work, Energy and Power

148933 Position of a $3 \mathrm{~kg}$ mass moving along the $x$-axis is given by $x=0.3 \cos (\omega t) \mathrm{m}$. If $K(t)$ denotes the kinetic energy at time $t$,
Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

1 $\frac{1}{3}$
2 $\frac{1}{2}$
3 $\frac{\sqrt{3}}{2}$
4 $\sqrt{3}$
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Work, Energy and Power

148930 If the kinetic energy of a body becomes four times then its momentum will be

1 $p_{\text {new }}=3 p_{\text {initial }}$
2 $\mathrm{p}_{\text {new }}=4 \mathrm{p}_{\text {initial }}$
3 $\mathrm{p}_{\text {new }}=2 \mathrm{p}_{\text {initial }}$
4 $p_{\text {new }}=p_{\text {initial }}$
Work, Energy and Power

148931 In the given figure, the block of mass $m$ is dropped from the point ' $A$ '. the expression for kinetic energy of block when it reaches point ' $B$ ' is

1 $\frac{1}{2} \mathrm{mgy}_{0}^{2}$
2 $\frac{1}{2} \mathrm{mgy}^{2}$
3 $\operatorname{mg}\left(\mathrm{y}-\mathrm{y}_{0}\right)$
4 $\mathrm{mgy}_{0}$
Work, Energy and Power

148932 Water falls from a $40 \mathrm{~m}$ high dam at the rate of $9 \times 10^{4} \mathrm{~kg}$ per hour. Fifty percent of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of $100 \mathrm{~W}$ lamps, that can be lit, is:
(Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

1 25
2 50
3 100
4 18
Work, Energy and Power

148933 Position of a $3 \mathrm{~kg}$ mass moving along the $x$-axis is given by $x=0.3 \cos (\omega t) \mathrm{m}$. If $K(t)$ denotes the kinetic energy at time $t$,
Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

1 $\frac{1}{3}$
2 $\frac{1}{2}$
3 $\frac{\sqrt{3}}{2}$
4 $\sqrt{3}$