149130
A cable in the form of a spiral roll (shown in the figure) has a linear density $\rho$. It is uncoiled at a uniform speed $v$, if the total length of the cable is $L$. The work done in uncoiling the cable is
1 $\rho L v^{2} / 4$
2 $\rho L v^{2} / 2$
3 $\rho \mathrm{Lv}^{2} / 3$
4 $\rho L v^{2}$
Explanation:
B Given that, Initial speed of the coil $(\mathrm{u})=0 \mathrm{~m} / \mathrm{s}$ Linear density $=\rho$ Final speed $=\mathrm{v}$ (uncoiling speed $)$ Length of the cable $=\mathrm{L}$ Mass of coil $=\rho \mathrm{L}$ So, Work done, $\mathrm{W}=\Delta \mathrm{KE}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$ $\mathrm{W}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$ $\mathrm{W}=\frac{1}{2} m v^{2}-0$ $\mathrm{W}=\frac{1}{2} \times(\rho \mathrm{L})(\mathrm{v})^{2}$ $\mathrm{W}=\frac{1}{2} \rho \mathrm{Lv}^{2}$
JCECE-2015
Work, Energy and Power
149131
A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively. The maximum velocity of this swing is:
1 $5 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $15 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Minimum height of swing from the earth's surface, $\mathrm{H}_{1}=0.75 \mathrm{~m}$ Maximum height of swing from earth's surface, $\mathrm{H}_{2}=2 \mathrm{~m}$ From energy conservation rule, Kinetic energy $=$ Potential energy $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}(2-0.75)$ $\mathrm{v}_{\text {max }}=\sqrt{2 \times 10 \times 1.25}$ $\mathrm{v}_{\text {max }}=\sqrt{25}$ $\mathrm{v}_{\text {max }}=5 \mathrm{~m} / \mathrm{s}$
JCECE-2004
Work, Energy and Power
149132
Two bodies of masses $0.1 \mathrm{~kg}$ and $0.4 \mathrm{~kg}$ move towards each other with the velocities $1 \mathrm{~m} / \mathrm{s}$ and $0.1 \mathrm{~m} / \mathrm{s}$ respectively. After collision they stick together. In 10 sec the combined mass travels:
1 $120 \mathrm{~m}$
2 $0.12 \mathrm{~m}$
3 $12 \mathrm{~m}$
4 $1.2 \mathrm{~m}$
Explanation:
D Given that, Mass of first body $\left(\mathrm{m}_{1}\right)=0.1 \mathrm{~kg}$ Mass of second body $\left(\mathrm{m}_{2}\right)=0.4 \mathrm{~kg}$ Velocity before collision $\left(\mathrm{v}_{1}\right)=1 \mathrm{~m} / \mathrm{s}$ Velocity before collision moving opposite direction $\left(\mathrm{v}_{2}\right)$ $=-0.1 \mathrm{~m} / \mathrm{s}$ $\mathrm{t}=10 \mathrm{~s}$ According to conservation of linear momentum, $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$ $(0.1)(1)+(0.4)(-0.1)=(0.1+0.4) \mathrm{v}$ $0.1-0.04=0.5 \mathrm{v}$ $\mathrm{v}=0.12 \mathrm{~m} / \mathrm{s}$ $\therefore$ Distance, $\mathrm{s}=\mathrm{vt}$ $\mathrm{s}=0.12 \times 10=1.2 \mathrm{~m}$
AIIMS-2010
Work, Energy and Power
149133
If the escape speed of a projectile on Earth's surface is $11.2 \mathrm{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.
1 $56.63 \mathrm{kms}^{-1}$
2 $33 \mathrm{kms}^{-1}$
3 $39 \mathrm{kms}^{-1}$
4 $31.7 \mathrm{kms}^{-1}$
Explanation:
D Given, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{kms}^{-1}$ Projected speed $\left(\mathrm{v}_{\mathrm{p}}\right)=3 \mathrm{v}_{\mathrm{e}}$ Mass of projectile $=\mathrm{m}$ Suppose energy of projectile far away from the earth $=$ $\mathrm{v}_{1}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Total Energy of the projectile far away from the earth $\left.=\frac{1}{2} \mathrm{mv}_{1}^{2} \quad \quad \text { (At infinite, P.E. }=0\right)$ Applying law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}$ $\mathrm{v}_{1}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{e}}^{2}}$ $\mathrm{v}_{1}=\sqrt{\left(3 \mathrm{v}_{\mathrm{e}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{1}=\sqrt{8} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{1}=\sqrt{8} \times 11.2$ $\mathrm{v}_{1}=31.6736 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{1} \simeq 31.7 \mathrm{kms}^{-1}$ Speed of the body at for away from the earth is $=31.7 \mathrm{kms}^{-1}$
BCECE-2015
Work, Energy and Power
149134
A spring of constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is-
149130
A cable in the form of a spiral roll (shown in the figure) has a linear density $\rho$. It is uncoiled at a uniform speed $v$, if the total length of the cable is $L$. The work done in uncoiling the cable is
1 $\rho L v^{2} / 4$
2 $\rho L v^{2} / 2$
3 $\rho \mathrm{Lv}^{2} / 3$
4 $\rho L v^{2}$
Explanation:
B Given that, Initial speed of the coil $(\mathrm{u})=0 \mathrm{~m} / \mathrm{s}$ Linear density $=\rho$ Final speed $=\mathrm{v}$ (uncoiling speed $)$ Length of the cable $=\mathrm{L}$ Mass of coil $=\rho \mathrm{L}$ So, Work done, $\mathrm{W}=\Delta \mathrm{KE}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$ $\mathrm{W}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$ $\mathrm{W}=\frac{1}{2} m v^{2}-0$ $\mathrm{W}=\frac{1}{2} \times(\rho \mathrm{L})(\mathrm{v})^{2}$ $\mathrm{W}=\frac{1}{2} \rho \mathrm{Lv}^{2}$
JCECE-2015
Work, Energy and Power
149131
A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively. The maximum velocity of this swing is:
1 $5 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $15 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Minimum height of swing from the earth's surface, $\mathrm{H}_{1}=0.75 \mathrm{~m}$ Maximum height of swing from earth's surface, $\mathrm{H}_{2}=2 \mathrm{~m}$ From energy conservation rule, Kinetic energy $=$ Potential energy $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}(2-0.75)$ $\mathrm{v}_{\text {max }}=\sqrt{2 \times 10 \times 1.25}$ $\mathrm{v}_{\text {max }}=\sqrt{25}$ $\mathrm{v}_{\text {max }}=5 \mathrm{~m} / \mathrm{s}$
JCECE-2004
Work, Energy and Power
149132
Two bodies of masses $0.1 \mathrm{~kg}$ and $0.4 \mathrm{~kg}$ move towards each other with the velocities $1 \mathrm{~m} / \mathrm{s}$ and $0.1 \mathrm{~m} / \mathrm{s}$ respectively. After collision they stick together. In 10 sec the combined mass travels:
1 $120 \mathrm{~m}$
2 $0.12 \mathrm{~m}$
3 $12 \mathrm{~m}$
4 $1.2 \mathrm{~m}$
Explanation:
D Given that, Mass of first body $\left(\mathrm{m}_{1}\right)=0.1 \mathrm{~kg}$ Mass of second body $\left(\mathrm{m}_{2}\right)=0.4 \mathrm{~kg}$ Velocity before collision $\left(\mathrm{v}_{1}\right)=1 \mathrm{~m} / \mathrm{s}$ Velocity before collision moving opposite direction $\left(\mathrm{v}_{2}\right)$ $=-0.1 \mathrm{~m} / \mathrm{s}$ $\mathrm{t}=10 \mathrm{~s}$ According to conservation of linear momentum, $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$ $(0.1)(1)+(0.4)(-0.1)=(0.1+0.4) \mathrm{v}$ $0.1-0.04=0.5 \mathrm{v}$ $\mathrm{v}=0.12 \mathrm{~m} / \mathrm{s}$ $\therefore$ Distance, $\mathrm{s}=\mathrm{vt}$ $\mathrm{s}=0.12 \times 10=1.2 \mathrm{~m}$
AIIMS-2010
Work, Energy and Power
149133
If the escape speed of a projectile on Earth's surface is $11.2 \mathrm{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.
1 $56.63 \mathrm{kms}^{-1}$
2 $33 \mathrm{kms}^{-1}$
3 $39 \mathrm{kms}^{-1}$
4 $31.7 \mathrm{kms}^{-1}$
Explanation:
D Given, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{kms}^{-1}$ Projected speed $\left(\mathrm{v}_{\mathrm{p}}\right)=3 \mathrm{v}_{\mathrm{e}}$ Mass of projectile $=\mathrm{m}$ Suppose energy of projectile far away from the earth $=$ $\mathrm{v}_{1}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Total Energy of the projectile far away from the earth $\left.=\frac{1}{2} \mathrm{mv}_{1}^{2} \quad \quad \text { (At infinite, P.E. }=0\right)$ Applying law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}$ $\mathrm{v}_{1}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{e}}^{2}}$ $\mathrm{v}_{1}=\sqrt{\left(3 \mathrm{v}_{\mathrm{e}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{1}=\sqrt{8} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{1}=\sqrt{8} \times 11.2$ $\mathrm{v}_{1}=31.6736 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{1} \simeq 31.7 \mathrm{kms}^{-1}$ Speed of the body at for away from the earth is $=31.7 \mathrm{kms}^{-1}$
BCECE-2015
Work, Energy and Power
149134
A spring of constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is-
149130
A cable in the form of a spiral roll (shown in the figure) has a linear density $\rho$. It is uncoiled at a uniform speed $v$, if the total length of the cable is $L$. The work done in uncoiling the cable is
1 $\rho L v^{2} / 4$
2 $\rho L v^{2} / 2$
3 $\rho \mathrm{Lv}^{2} / 3$
4 $\rho L v^{2}$
Explanation:
B Given that, Initial speed of the coil $(\mathrm{u})=0 \mathrm{~m} / \mathrm{s}$ Linear density $=\rho$ Final speed $=\mathrm{v}$ (uncoiling speed $)$ Length of the cable $=\mathrm{L}$ Mass of coil $=\rho \mathrm{L}$ So, Work done, $\mathrm{W}=\Delta \mathrm{KE}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$ $\mathrm{W}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$ $\mathrm{W}=\frac{1}{2} m v^{2}-0$ $\mathrm{W}=\frac{1}{2} \times(\rho \mathrm{L})(\mathrm{v})^{2}$ $\mathrm{W}=\frac{1}{2} \rho \mathrm{Lv}^{2}$
JCECE-2015
Work, Energy and Power
149131
A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively. The maximum velocity of this swing is:
1 $5 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $15 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Minimum height of swing from the earth's surface, $\mathrm{H}_{1}=0.75 \mathrm{~m}$ Maximum height of swing from earth's surface, $\mathrm{H}_{2}=2 \mathrm{~m}$ From energy conservation rule, Kinetic energy $=$ Potential energy $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}(2-0.75)$ $\mathrm{v}_{\text {max }}=\sqrt{2 \times 10 \times 1.25}$ $\mathrm{v}_{\text {max }}=\sqrt{25}$ $\mathrm{v}_{\text {max }}=5 \mathrm{~m} / \mathrm{s}$
JCECE-2004
Work, Energy and Power
149132
Two bodies of masses $0.1 \mathrm{~kg}$ and $0.4 \mathrm{~kg}$ move towards each other with the velocities $1 \mathrm{~m} / \mathrm{s}$ and $0.1 \mathrm{~m} / \mathrm{s}$ respectively. After collision they stick together. In 10 sec the combined mass travels:
1 $120 \mathrm{~m}$
2 $0.12 \mathrm{~m}$
3 $12 \mathrm{~m}$
4 $1.2 \mathrm{~m}$
Explanation:
D Given that, Mass of first body $\left(\mathrm{m}_{1}\right)=0.1 \mathrm{~kg}$ Mass of second body $\left(\mathrm{m}_{2}\right)=0.4 \mathrm{~kg}$ Velocity before collision $\left(\mathrm{v}_{1}\right)=1 \mathrm{~m} / \mathrm{s}$ Velocity before collision moving opposite direction $\left(\mathrm{v}_{2}\right)$ $=-0.1 \mathrm{~m} / \mathrm{s}$ $\mathrm{t}=10 \mathrm{~s}$ According to conservation of linear momentum, $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$ $(0.1)(1)+(0.4)(-0.1)=(0.1+0.4) \mathrm{v}$ $0.1-0.04=0.5 \mathrm{v}$ $\mathrm{v}=0.12 \mathrm{~m} / \mathrm{s}$ $\therefore$ Distance, $\mathrm{s}=\mathrm{vt}$ $\mathrm{s}=0.12 \times 10=1.2 \mathrm{~m}$
AIIMS-2010
Work, Energy and Power
149133
If the escape speed of a projectile on Earth's surface is $11.2 \mathrm{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.
1 $56.63 \mathrm{kms}^{-1}$
2 $33 \mathrm{kms}^{-1}$
3 $39 \mathrm{kms}^{-1}$
4 $31.7 \mathrm{kms}^{-1}$
Explanation:
D Given, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{kms}^{-1}$ Projected speed $\left(\mathrm{v}_{\mathrm{p}}\right)=3 \mathrm{v}_{\mathrm{e}}$ Mass of projectile $=\mathrm{m}$ Suppose energy of projectile far away from the earth $=$ $\mathrm{v}_{1}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Total Energy of the projectile far away from the earth $\left.=\frac{1}{2} \mathrm{mv}_{1}^{2} \quad \quad \text { (At infinite, P.E. }=0\right)$ Applying law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}$ $\mathrm{v}_{1}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{e}}^{2}}$ $\mathrm{v}_{1}=\sqrt{\left(3 \mathrm{v}_{\mathrm{e}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{1}=\sqrt{8} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{1}=\sqrt{8} \times 11.2$ $\mathrm{v}_{1}=31.6736 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{1} \simeq 31.7 \mathrm{kms}^{-1}$ Speed of the body at for away from the earth is $=31.7 \mathrm{kms}^{-1}$
BCECE-2015
Work, Energy and Power
149134
A spring of constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is-
NEET Test Series from KOTA - 10 Papers In MS WORD
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Work, Energy and Power
149130
A cable in the form of a spiral roll (shown in the figure) has a linear density $\rho$. It is uncoiled at a uniform speed $v$, if the total length of the cable is $L$. The work done in uncoiling the cable is
1 $\rho L v^{2} / 4$
2 $\rho L v^{2} / 2$
3 $\rho \mathrm{Lv}^{2} / 3$
4 $\rho L v^{2}$
Explanation:
B Given that, Initial speed of the coil $(\mathrm{u})=0 \mathrm{~m} / \mathrm{s}$ Linear density $=\rho$ Final speed $=\mathrm{v}$ (uncoiling speed $)$ Length of the cable $=\mathrm{L}$ Mass of coil $=\rho \mathrm{L}$ So, Work done, $\mathrm{W}=\Delta \mathrm{KE}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$ $\mathrm{W}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$ $\mathrm{W}=\frac{1}{2} m v^{2}-0$ $\mathrm{W}=\frac{1}{2} \times(\rho \mathrm{L})(\mathrm{v})^{2}$ $\mathrm{W}=\frac{1}{2} \rho \mathrm{Lv}^{2}$
JCECE-2015
Work, Energy and Power
149131
A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively. The maximum velocity of this swing is:
1 $5 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $15 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Minimum height of swing from the earth's surface, $\mathrm{H}_{1}=0.75 \mathrm{~m}$ Maximum height of swing from earth's surface, $\mathrm{H}_{2}=2 \mathrm{~m}$ From energy conservation rule, Kinetic energy $=$ Potential energy $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}(2-0.75)$ $\mathrm{v}_{\text {max }}=\sqrt{2 \times 10 \times 1.25}$ $\mathrm{v}_{\text {max }}=\sqrt{25}$ $\mathrm{v}_{\text {max }}=5 \mathrm{~m} / \mathrm{s}$
JCECE-2004
Work, Energy and Power
149132
Two bodies of masses $0.1 \mathrm{~kg}$ and $0.4 \mathrm{~kg}$ move towards each other with the velocities $1 \mathrm{~m} / \mathrm{s}$ and $0.1 \mathrm{~m} / \mathrm{s}$ respectively. After collision they stick together. In 10 sec the combined mass travels:
1 $120 \mathrm{~m}$
2 $0.12 \mathrm{~m}$
3 $12 \mathrm{~m}$
4 $1.2 \mathrm{~m}$
Explanation:
D Given that, Mass of first body $\left(\mathrm{m}_{1}\right)=0.1 \mathrm{~kg}$ Mass of second body $\left(\mathrm{m}_{2}\right)=0.4 \mathrm{~kg}$ Velocity before collision $\left(\mathrm{v}_{1}\right)=1 \mathrm{~m} / \mathrm{s}$ Velocity before collision moving opposite direction $\left(\mathrm{v}_{2}\right)$ $=-0.1 \mathrm{~m} / \mathrm{s}$ $\mathrm{t}=10 \mathrm{~s}$ According to conservation of linear momentum, $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$ $(0.1)(1)+(0.4)(-0.1)=(0.1+0.4) \mathrm{v}$ $0.1-0.04=0.5 \mathrm{v}$ $\mathrm{v}=0.12 \mathrm{~m} / \mathrm{s}$ $\therefore$ Distance, $\mathrm{s}=\mathrm{vt}$ $\mathrm{s}=0.12 \times 10=1.2 \mathrm{~m}$
AIIMS-2010
Work, Energy and Power
149133
If the escape speed of a projectile on Earth's surface is $11.2 \mathrm{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.
1 $56.63 \mathrm{kms}^{-1}$
2 $33 \mathrm{kms}^{-1}$
3 $39 \mathrm{kms}^{-1}$
4 $31.7 \mathrm{kms}^{-1}$
Explanation:
D Given, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{kms}^{-1}$ Projected speed $\left(\mathrm{v}_{\mathrm{p}}\right)=3 \mathrm{v}_{\mathrm{e}}$ Mass of projectile $=\mathrm{m}$ Suppose energy of projectile far away from the earth $=$ $\mathrm{v}_{1}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Total Energy of the projectile far away from the earth $\left.=\frac{1}{2} \mathrm{mv}_{1}^{2} \quad \quad \text { (At infinite, P.E. }=0\right)$ Applying law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}$ $\mathrm{v}_{1}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{e}}^{2}}$ $\mathrm{v}_{1}=\sqrt{\left(3 \mathrm{v}_{\mathrm{e}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{1}=\sqrt{8} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{1}=\sqrt{8} \times 11.2$ $\mathrm{v}_{1}=31.6736 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{1} \simeq 31.7 \mathrm{kms}^{-1}$ Speed of the body at for away from the earth is $=31.7 \mathrm{kms}^{-1}$
BCECE-2015
Work, Energy and Power
149134
A spring of constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is-
149130
A cable in the form of a spiral roll (shown in the figure) has a linear density $\rho$. It is uncoiled at a uniform speed $v$, if the total length of the cable is $L$. The work done in uncoiling the cable is
1 $\rho L v^{2} / 4$
2 $\rho L v^{2} / 2$
3 $\rho \mathrm{Lv}^{2} / 3$
4 $\rho L v^{2}$
Explanation:
B Given that, Initial speed of the coil $(\mathrm{u})=0 \mathrm{~m} / \mathrm{s}$ Linear density $=\rho$ Final speed $=\mathrm{v}$ (uncoiling speed $)$ Length of the cable $=\mathrm{L}$ Mass of coil $=\rho \mathrm{L}$ So, Work done, $\mathrm{W}=\Delta \mathrm{KE}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$ $\mathrm{W}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$ $\mathrm{W}=\frac{1}{2} m v^{2}-0$ $\mathrm{W}=\frac{1}{2} \times(\rho \mathrm{L})(\mathrm{v})^{2}$ $\mathrm{W}=\frac{1}{2} \rho \mathrm{Lv}^{2}$
JCECE-2015
Work, Energy and Power
149131
A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 \mathrm{~m}$ and $2 \mathrm{~m}$ respectively. The maximum velocity of this swing is:
1 $5 \mathrm{~m} / \mathrm{s}$
2 $10 \mathrm{~m} / \mathrm{s}$
3 $15 \mathrm{~m} / \mathrm{s}$
4 $20 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given that, Minimum height of swing from the earth's surface, $\mathrm{H}_{1}=0.75 \mathrm{~m}$ Maximum height of swing from earth's surface, $\mathrm{H}_{2}=2 \mathrm{~m}$ From energy conservation rule, Kinetic energy $=$ Potential energy $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{mg}(2-0.75)$ $\mathrm{v}_{\text {max }}=\sqrt{2 \times 10 \times 1.25}$ $\mathrm{v}_{\text {max }}=\sqrt{25}$ $\mathrm{v}_{\text {max }}=5 \mathrm{~m} / \mathrm{s}$
JCECE-2004
Work, Energy and Power
149132
Two bodies of masses $0.1 \mathrm{~kg}$ and $0.4 \mathrm{~kg}$ move towards each other with the velocities $1 \mathrm{~m} / \mathrm{s}$ and $0.1 \mathrm{~m} / \mathrm{s}$ respectively. After collision they stick together. In 10 sec the combined mass travels:
1 $120 \mathrm{~m}$
2 $0.12 \mathrm{~m}$
3 $12 \mathrm{~m}$
4 $1.2 \mathrm{~m}$
Explanation:
D Given that, Mass of first body $\left(\mathrm{m}_{1}\right)=0.1 \mathrm{~kg}$ Mass of second body $\left(\mathrm{m}_{2}\right)=0.4 \mathrm{~kg}$ Velocity before collision $\left(\mathrm{v}_{1}\right)=1 \mathrm{~m} / \mathrm{s}$ Velocity before collision moving opposite direction $\left(\mathrm{v}_{2}\right)$ $=-0.1 \mathrm{~m} / \mathrm{s}$ $\mathrm{t}=10 \mathrm{~s}$ According to conservation of linear momentum, $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$ $(0.1)(1)+(0.4)(-0.1)=(0.1+0.4) \mathrm{v}$ $0.1-0.04=0.5 \mathrm{v}$ $\mathrm{v}=0.12 \mathrm{~m} / \mathrm{s}$ $\therefore$ Distance, $\mathrm{s}=\mathrm{vt}$ $\mathrm{s}=0.12 \times 10=1.2 \mathrm{~m}$
AIIMS-2010
Work, Energy and Power
149133
If the escape speed of a projectile on Earth's surface is $11.2 \mathrm{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.
1 $56.63 \mathrm{kms}^{-1}$
2 $33 \mathrm{kms}^{-1}$
3 $39 \mathrm{kms}^{-1}$
4 $31.7 \mathrm{kms}^{-1}$
Explanation:
D Given, Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{kms}^{-1}$ Projected speed $\left(\mathrm{v}_{\mathrm{p}}\right)=3 \mathrm{v}_{\mathrm{e}}$ Mass of projectile $=\mathrm{m}$ Suppose energy of projectile far away from the earth $=$ $\mathrm{v}_{1}$ $\text { Total energy }=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ Total Energy of the projectile far away from the earth $\left.=\frac{1}{2} \mathrm{mv}_{1}^{2} \quad \quad \text { (At infinite, P.E. }=0\right)$ Applying law of conservation of energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}$ $\mathrm{v}_{1}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{e}}^{2}}$ $\mathrm{v}_{1}=\sqrt{\left(3 \mathrm{v}_{\mathrm{e}}\right)^{2}-\left(\mathrm{v}_{\mathrm{e}}\right)^{2}}$ $\mathrm{v}_{1}=\sqrt{8} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{1}=\sqrt{8} \times 11.2$ $\mathrm{v}_{1}=31.6736 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{1} \simeq 31.7 \mathrm{kms}^{-1}$ Speed of the body at for away from the earth is $=31.7 \mathrm{kms}^{-1}$
BCECE-2015
Work, Energy and Power
149134
A spring of constant $5 \times 10^{3} \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is-
