NEET Test Series from KOTA - 10 Papers In MS WORD
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Work, Energy and Power
148921
From a building two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then:
B Given, two balls A and B are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). From law of conservation of energy, Potential energy at height $(\mathrm{h}) \stackrel{=}{=}$ Kinetic energy at ground Since, $\quad \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\begin{array}{ll} \mathrm{gh}=\frac{1}{2} \mathrm{v}^{2} \\ \mathrm{v}^{2}=2 \mathrm{gh} \\ \mathrm{v} =\sqrt{2 \mathrm{gh}} \\ \therefore \quad \mathrm{v} \propto \sqrt{\mathrm{h}} \\ \text { Then, } \quad \mathrm{v}_{\mathrm{A}} \propto \sqrt{\mathrm{h}} \\ \text { And, } \quad \mathrm{v}_{\mathrm{B}} \propto \sqrt{\mathrm{h}} \\ \text { Therefore, } \quad \mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}\end{array}$
AMU-2015
Work, Energy and Power
148922
A mass of $\mathbf{4 0 0} \mathrm{g}$ and a mass of $100 \mathrm{~g}$ have same $\mathrm{KE}$, then the ratio of their momentums will be
148923
Two bodies $A$ and $B$ having masses in the ratio of 3:1 possess the same kinetic energy. The ratio of linear momentum of $B$ to $A$ is
1 $1: 3$
2 $3: 1$
3 $1: \sqrt{3}$
4 $\sqrt{3}: 1$
Explanation:
C Given Ratio of mass of to bodies $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}=3: 1$ Let, $\text { Momentum of }(A)=p_{A}$ $\text { Momentum of }(B)=p_{B}$ We know that, relation between kinetic energy and momentum. $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ $\mathrm{p}_{\mathrm{A}}=\sqrt{2 \mathrm{~m}_{\mathrm{A}}(\mathrm{KE})_{\mathrm{A}}}$ $\mathrm{p}_{\mathrm{B}}=\sqrt{2 \mathrm{~m}_{\mathrm{B}}(\mathrm{KE})_{\mathrm{B}}}$ Divide equation (ii) by equation (i), we get- $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{B}}}}{\sqrt{\mathrm{m}_{\mathrm{A}}}}$ ${\left[\mathrm{m}_{\mathrm{A}}: \mathrm{m}_{\mathrm{B}}=3: 1 \text { and }(\mathrm{KE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}\right]}$ $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ So, ratio of linear momentum of $\mathrm{B}$ to $\mathrm{A}$ is $\mathrm{p}_{\mathrm{B}}: \mathrm{p}_{\mathrm{A}}=1: \sqrt{3}$
VITEEE-2017
Work, Energy and Power
148924
The kinetic energy of a body becomes four times its initial value. The new linear momentum will be:
1 eight times of the initial value
2 four times of the initial value
3 twice of the initial value
4 remain as the initial value
Explanation:
C Given that, Kinetic energy of a body becomes four times its initial value. Let, initial kinetic energy $=$ K.E. $\therefore$ Final kinetic energy $=4$ K.E. We know that, $\text { K.E. }=\frac{p_{1}^{2}}{2 m}$ and 4 K.E. $=\frac{p_{2}^{2}}{2 m}$ Dividing equation (i) by equation (ii), we get- $\frac{\text { K.E. }}{4 \text { K.E. }}=\frac{\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}}}$ $\frac{\text { K.E. }}{4 \mathrm{~K} . \mathrm{E} .}=\frac{\mathrm{p}_{1}^{2}}{\mathrm{p}_{2}^{2}}$ $\mathrm{p}_{2}^{2}=4 \mathrm{p}_{1}^{2}$ $\mathrm{p}_{2}=\sqrt{4 \mathrm{p}_{1}^{2}}$ $\mathrm{p}_{2}=2 \mathrm{p}_{1}$ So, the new linear momentum will be twice of the initial value.
148921
From a building two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then:
B Given, two balls A and B are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). From law of conservation of energy, Potential energy at height $(\mathrm{h}) \stackrel{=}{=}$ Kinetic energy at ground Since, $\quad \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\begin{array}{ll} \mathrm{gh}=\frac{1}{2} \mathrm{v}^{2} \\ \mathrm{v}^{2}=2 \mathrm{gh} \\ \mathrm{v} =\sqrt{2 \mathrm{gh}} \\ \therefore \quad \mathrm{v} \propto \sqrt{\mathrm{h}} \\ \text { Then, } \quad \mathrm{v}_{\mathrm{A}} \propto \sqrt{\mathrm{h}} \\ \text { And, } \quad \mathrm{v}_{\mathrm{B}} \propto \sqrt{\mathrm{h}} \\ \text { Therefore, } \quad \mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}\end{array}$
AMU-2015
Work, Energy and Power
148922
A mass of $\mathbf{4 0 0} \mathrm{g}$ and a mass of $100 \mathrm{~g}$ have same $\mathrm{KE}$, then the ratio of their momentums will be
148923
Two bodies $A$ and $B$ having masses in the ratio of 3:1 possess the same kinetic energy. The ratio of linear momentum of $B$ to $A$ is
1 $1: 3$
2 $3: 1$
3 $1: \sqrt{3}$
4 $\sqrt{3}: 1$
Explanation:
C Given Ratio of mass of to bodies $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}=3: 1$ Let, $\text { Momentum of }(A)=p_{A}$ $\text { Momentum of }(B)=p_{B}$ We know that, relation between kinetic energy and momentum. $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ $\mathrm{p}_{\mathrm{A}}=\sqrt{2 \mathrm{~m}_{\mathrm{A}}(\mathrm{KE})_{\mathrm{A}}}$ $\mathrm{p}_{\mathrm{B}}=\sqrt{2 \mathrm{~m}_{\mathrm{B}}(\mathrm{KE})_{\mathrm{B}}}$ Divide equation (ii) by equation (i), we get- $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{B}}}}{\sqrt{\mathrm{m}_{\mathrm{A}}}}$ ${\left[\mathrm{m}_{\mathrm{A}}: \mathrm{m}_{\mathrm{B}}=3: 1 \text { and }(\mathrm{KE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}\right]}$ $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ So, ratio of linear momentum of $\mathrm{B}$ to $\mathrm{A}$ is $\mathrm{p}_{\mathrm{B}}: \mathrm{p}_{\mathrm{A}}=1: \sqrt{3}$
VITEEE-2017
Work, Energy and Power
148924
The kinetic energy of a body becomes four times its initial value. The new linear momentum will be:
1 eight times of the initial value
2 four times of the initial value
3 twice of the initial value
4 remain as the initial value
Explanation:
C Given that, Kinetic energy of a body becomes four times its initial value. Let, initial kinetic energy $=$ K.E. $\therefore$ Final kinetic energy $=4$ K.E. We know that, $\text { K.E. }=\frac{p_{1}^{2}}{2 m}$ and 4 K.E. $=\frac{p_{2}^{2}}{2 m}$ Dividing equation (i) by equation (ii), we get- $\frac{\text { K.E. }}{4 \text { K.E. }}=\frac{\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}}}$ $\frac{\text { K.E. }}{4 \mathrm{~K} . \mathrm{E} .}=\frac{\mathrm{p}_{1}^{2}}{\mathrm{p}_{2}^{2}}$ $\mathrm{p}_{2}^{2}=4 \mathrm{p}_{1}^{2}$ $\mathrm{p}_{2}=\sqrt{4 \mathrm{p}_{1}^{2}}$ $\mathrm{p}_{2}=2 \mathrm{p}_{1}$ So, the new linear momentum will be twice of the initial value.
148921
From a building two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then:
B Given, two balls A and B are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). From law of conservation of energy, Potential energy at height $(\mathrm{h}) \stackrel{=}{=}$ Kinetic energy at ground Since, $\quad \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\begin{array}{ll} \mathrm{gh}=\frac{1}{2} \mathrm{v}^{2} \\ \mathrm{v}^{2}=2 \mathrm{gh} \\ \mathrm{v} =\sqrt{2 \mathrm{gh}} \\ \therefore \quad \mathrm{v} \propto \sqrt{\mathrm{h}} \\ \text { Then, } \quad \mathrm{v}_{\mathrm{A}} \propto \sqrt{\mathrm{h}} \\ \text { And, } \quad \mathrm{v}_{\mathrm{B}} \propto \sqrt{\mathrm{h}} \\ \text { Therefore, } \quad \mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}\end{array}$
AMU-2015
Work, Energy and Power
148922
A mass of $\mathbf{4 0 0} \mathrm{g}$ and a mass of $100 \mathrm{~g}$ have same $\mathrm{KE}$, then the ratio of their momentums will be
148923
Two bodies $A$ and $B$ having masses in the ratio of 3:1 possess the same kinetic energy. The ratio of linear momentum of $B$ to $A$ is
1 $1: 3$
2 $3: 1$
3 $1: \sqrt{3}$
4 $\sqrt{3}: 1$
Explanation:
C Given Ratio of mass of to bodies $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}=3: 1$ Let, $\text { Momentum of }(A)=p_{A}$ $\text { Momentum of }(B)=p_{B}$ We know that, relation between kinetic energy and momentum. $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ $\mathrm{p}_{\mathrm{A}}=\sqrt{2 \mathrm{~m}_{\mathrm{A}}(\mathrm{KE})_{\mathrm{A}}}$ $\mathrm{p}_{\mathrm{B}}=\sqrt{2 \mathrm{~m}_{\mathrm{B}}(\mathrm{KE})_{\mathrm{B}}}$ Divide equation (ii) by equation (i), we get- $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{B}}}}{\sqrt{\mathrm{m}_{\mathrm{A}}}}$ ${\left[\mathrm{m}_{\mathrm{A}}: \mathrm{m}_{\mathrm{B}}=3: 1 \text { and }(\mathrm{KE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}\right]}$ $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ So, ratio of linear momentum of $\mathrm{B}$ to $\mathrm{A}$ is $\mathrm{p}_{\mathrm{B}}: \mathrm{p}_{\mathrm{A}}=1: \sqrt{3}$
VITEEE-2017
Work, Energy and Power
148924
The kinetic energy of a body becomes four times its initial value. The new linear momentum will be:
1 eight times of the initial value
2 four times of the initial value
3 twice of the initial value
4 remain as the initial value
Explanation:
C Given that, Kinetic energy of a body becomes four times its initial value. Let, initial kinetic energy $=$ K.E. $\therefore$ Final kinetic energy $=4$ K.E. We know that, $\text { K.E. }=\frac{p_{1}^{2}}{2 m}$ and 4 K.E. $=\frac{p_{2}^{2}}{2 m}$ Dividing equation (i) by equation (ii), we get- $\frac{\text { K.E. }}{4 \text { K.E. }}=\frac{\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}}}$ $\frac{\text { K.E. }}{4 \mathrm{~K} . \mathrm{E} .}=\frac{\mathrm{p}_{1}^{2}}{\mathrm{p}_{2}^{2}}$ $\mathrm{p}_{2}^{2}=4 \mathrm{p}_{1}^{2}$ $\mathrm{p}_{2}=\sqrt{4 \mathrm{p}_{1}^{2}}$ $\mathrm{p}_{2}=2 \mathrm{p}_{1}$ So, the new linear momentum will be twice of the initial value.
148921
From a building two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then:
B Given, two balls A and B are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). From law of conservation of energy, Potential energy at height $(\mathrm{h}) \stackrel{=}{=}$ Kinetic energy at ground Since, $\quad \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}$ $\begin{array}{ll} \mathrm{gh}=\frac{1}{2} \mathrm{v}^{2} \\ \mathrm{v}^{2}=2 \mathrm{gh} \\ \mathrm{v} =\sqrt{2 \mathrm{gh}} \\ \therefore \quad \mathrm{v} \propto \sqrt{\mathrm{h}} \\ \text { Then, } \quad \mathrm{v}_{\mathrm{A}} \propto \sqrt{\mathrm{h}} \\ \text { And, } \quad \mathrm{v}_{\mathrm{B}} \propto \sqrt{\mathrm{h}} \\ \text { Therefore, } \quad \mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}\end{array}$
AMU-2015
Work, Energy and Power
148922
A mass of $\mathbf{4 0 0} \mathrm{g}$ and a mass of $100 \mathrm{~g}$ have same $\mathrm{KE}$, then the ratio of their momentums will be
148923
Two bodies $A$ and $B$ having masses in the ratio of 3:1 possess the same kinetic energy. The ratio of linear momentum of $B$ to $A$ is
1 $1: 3$
2 $3: 1$
3 $1: \sqrt{3}$
4 $\sqrt{3}: 1$
Explanation:
C Given Ratio of mass of to bodies $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}=3: 1$ Let, $\text { Momentum of }(A)=p_{A}$ $\text { Momentum of }(B)=p_{B}$ We know that, relation between kinetic energy and momentum. $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ $\mathrm{p}_{\mathrm{A}}=\sqrt{2 \mathrm{~m}_{\mathrm{A}}(\mathrm{KE})_{\mathrm{A}}}$ $\mathrm{p}_{\mathrm{B}}=\sqrt{2 \mathrm{~m}_{\mathrm{B}}(\mathrm{KE})_{\mathrm{B}}}$ Divide equation (ii) by equation (i), we get- $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{m}_{\mathrm{B}}}}{\sqrt{\mathrm{m}_{\mathrm{A}}}}$ ${\left[\mathrm{m}_{\mathrm{A}}: \mathrm{m}_{\mathrm{B}}=3: 1 \text { and }(\mathrm{KE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}\right]}$ $\frac{\mathrm{p}_{\mathrm{B}}}{\mathrm{p}_{\mathrm{A}}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ So, ratio of linear momentum of $\mathrm{B}$ to $\mathrm{A}$ is $\mathrm{p}_{\mathrm{B}}: \mathrm{p}_{\mathrm{A}}=1: \sqrt{3}$
VITEEE-2017
Work, Energy and Power
148924
The kinetic energy of a body becomes four times its initial value. The new linear momentum will be:
1 eight times of the initial value
2 four times of the initial value
3 twice of the initial value
4 remain as the initial value
Explanation:
C Given that, Kinetic energy of a body becomes four times its initial value. Let, initial kinetic energy $=$ K.E. $\therefore$ Final kinetic energy $=4$ K.E. We know that, $\text { K.E. }=\frac{p_{1}^{2}}{2 m}$ and 4 K.E. $=\frac{p_{2}^{2}}{2 m}$ Dividing equation (i) by equation (ii), we get- $\frac{\text { K.E. }}{4 \text { K.E. }}=\frac{\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}}}$ $\frac{\text { K.E. }}{4 \mathrm{~K} . \mathrm{E} .}=\frac{\mathrm{p}_{1}^{2}}{\mathrm{p}_{2}^{2}}$ $\mathrm{p}_{2}^{2}=4 \mathrm{p}_{1}^{2}$ $\mathrm{p}_{2}=\sqrt{4 \mathrm{p}_{1}^{2}}$ $\mathrm{p}_{2}=2 \mathrm{p}_{1}$ So, the new linear momentum will be twice of the initial value.
