NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138336
The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $\rho_{1}$ and $\rho_{2}$. The ratio of the accelerations due to gravity at their surfaces is
D Given, radii of two planets is $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ Density of two planets is $\rho_{1}$ and $\rho_{2}$ For planet first $\mathrm{g}_{1}=\frac{\mathrm{Gm}_{1}}{\left(\mathrm{R}_{1}\right)^{2}}=\frac{\mathrm{G} \frac{4}{3} \pi\left(\mathrm{R}_{1}\right)^{3} \cdot \rho_{1}}{\left(\mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{1}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{1} \cdot \rho_{1}$ Similarly for second planet $\mathrm{g}_{2}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{2} \cdot \rho_{2}$ On dividing eq ${ }^{\mathrm{n}}$ (i) and (ii) we get $\frac{g_{1}}{g_{2}}=\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}$ $g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}$
MP PET 1994
Gravitation
138337
If the earth were to cease rotating about its own axis. The increase in the value of $g$ in C.G.S. system at a place of latitude of $45^{\circ}$ will be
1 2.68
2 1.68
3 3.36
4 0.34
Explanation:
B Given $\lambda=45^{\circ}, \mathrm{R}=6400 \times 10^{3} \mathrm{~m}$ $\because \quad \omega=\frac{2 \pi}{\mathrm{T}}$ $\omega=\frac{2 \pi}{24 \times 60 \times 60} \quad(\because \mathrm{T}=24 \mathrm{hr})$ The value of acceleration due to gravity with latitude $\lambda$ due to rotation of earth is, $g^{\prime}=g-R \omega^{2} \cos ^{2} \lambda$ $g^{\prime}-g=-R \omega^{2} \cos ^{2} \lambda$ $g-g^{\prime}= 6400 \times 10^{3} \times\left(\frac{2 \times 3.14}{24 \times 60 \times 60}\right)^{2} \times\left(\cos ^{2} 45^{\circ}\right)$ $g-g^{\prime}= \frac{6400 \times 10^{3} \times 4 \times 3.14 \times 3.14}{24 \times 60 \times 60 \times 24 \times 60 \times 60} \times \frac{1}{2}$ $= 16.89 \times 10^{-3} \mathrm{~m} / \mathrm{sec}^{2}$ $= 16.89 \times 10^{-1} \mathrm{~cm} / \mathrm{sec}^{2}$ $= 1.68 \mathrm{~cm} / \mathrm{sec}^{2}$
AIIMS-2010
Gravitation
138338
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is about $3200 \mathrm{~km}$. The mass of the earth is about 10 times of the mass. The object weighs $200 \mathrm{~N}$ on earth surface, then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Mass of earth $\left(M_{e}\right)=10$ time of object Weight of the object on earth $\left(\mathrm{W}_{\mathrm{e}}\right)=200 \mathrm{~N}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{Mg}_{\mathrm{m}}}{\mathrm{Mg}_{\mathrm{e}}}=\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} \times\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2} \times \mathrm{GM}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\mathrm{M}_{\mathrm{e}}\left(\mathrm{R}_{\mathrm{m}}\right)^{2}} \quad\left\{\because \mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}\right\}$ $=\frac{1}{10} \times(2)^{2} \quad\left[\because \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}=\frac{6400}{3200}=2\right]$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{4}{10}=\frac{2}{5} \quad \mathrm{~W}_{\mathrm{m}}=\mathrm{W}_{\mathrm{e}} \times \frac{2}{5}$ $\mathrm{~W}_{\mathrm{m}}=200 \times \frac{2}{5}=80 \mathrm{~N}$
AIIMS-2002
Gravitation
138335
Assertion: In a free fall, weight of a body becomes effectively zero. Reason: Acceleration due to gravity acting on a body having free fall is zero.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C In a free fall effective weight of a body becomes zero. The reason behind it, is the normal force acting on the object is equal to $\mathrm{mg}$ which is same as the weight of the object. So, the object feels weightless. So, Assertion is correct but reason is false.
138336
The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $\rho_{1}$ and $\rho_{2}$. The ratio of the accelerations due to gravity at their surfaces is
D Given, radii of two planets is $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ Density of two planets is $\rho_{1}$ and $\rho_{2}$ For planet first $\mathrm{g}_{1}=\frac{\mathrm{Gm}_{1}}{\left(\mathrm{R}_{1}\right)^{2}}=\frac{\mathrm{G} \frac{4}{3} \pi\left(\mathrm{R}_{1}\right)^{3} \cdot \rho_{1}}{\left(\mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{1}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{1} \cdot \rho_{1}$ Similarly for second planet $\mathrm{g}_{2}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{2} \cdot \rho_{2}$ On dividing eq ${ }^{\mathrm{n}}$ (i) and (ii) we get $\frac{g_{1}}{g_{2}}=\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}$ $g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}$
MP PET 1994
Gravitation
138337
If the earth were to cease rotating about its own axis. The increase in the value of $g$ in C.G.S. system at a place of latitude of $45^{\circ}$ will be
1 2.68
2 1.68
3 3.36
4 0.34
Explanation:
B Given $\lambda=45^{\circ}, \mathrm{R}=6400 \times 10^{3} \mathrm{~m}$ $\because \quad \omega=\frac{2 \pi}{\mathrm{T}}$ $\omega=\frac{2 \pi}{24 \times 60 \times 60} \quad(\because \mathrm{T}=24 \mathrm{hr})$ The value of acceleration due to gravity with latitude $\lambda$ due to rotation of earth is, $g^{\prime}=g-R \omega^{2} \cos ^{2} \lambda$ $g^{\prime}-g=-R \omega^{2} \cos ^{2} \lambda$ $g-g^{\prime}= 6400 \times 10^{3} \times\left(\frac{2 \times 3.14}{24 \times 60 \times 60}\right)^{2} \times\left(\cos ^{2} 45^{\circ}\right)$ $g-g^{\prime}= \frac{6400 \times 10^{3} \times 4 \times 3.14 \times 3.14}{24 \times 60 \times 60 \times 24 \times 60 \times 60} \times \frac{1}{2}$ $= 16.89 \times 10^{-3} \mathrm{~m} / \mathrm{sec}^{2}$ $= 16.89 \times 10^{-1} \mathrm{~cm} / \mathrm{sec}^{2}$ $= 1.68 \mathrm{~cm} / \mathrm{sec}^{2}$
AIIMS-2010
Gravitation
138338
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is about $3200 \mathrm{~km}$. The mass of the earth is about 10 times of the mass. The object weighs $200 \mathrm{~N}$ on earth surface, then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Mass of earth $\left(M_{e}\right)=10$ time of object Weight of the object on earth $\left(\mathrm{W}_{\mathrm{e}}\right)=200 \mathrm{~N}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{Mg}_{\mathrm{m}}}{\mathrm{Mg}_{\mathrm{e}}}=\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} \times\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2} \times \mathrm{GM}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\mathrm{M}_{\mathrm{e}}\left(\mathrm{R}_{\mathrm{m}}\right)^{2}} \quad\left\{\because \mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}\right\}$ $=\frac{1}{10} \times(2)^{2} \quad\left[\because \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}=\frac{6400}{3200}=2\right]$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{4}{10}=\frac{2}{5} \quad \mathrm{~W}_{\mathrm{m}}=\mathrm{W}_{\mathrm{e}} \times \frac{2}{5}$ $\mathrm{~W}_{\mathrm{m}}=200 \times \frac{2}{5}=80 \mathrm{~N}$
AIIMS-2002
Gravitation
138335
Assertion: In a free fall, weight of a body becomes effectively zero. Reason: Acceleration due to gravity acting on a body having free fall is zero.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C In a free fall effective weight of a body becomes zero. The reason behind it, is the normal force acting on the object is equal to $\mathrm{mg}$ which is same as the weight of the object. So, the object feels weightless. So, Assertion is correct but reason is false.
138336
The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $\rho_{1}$ and $\rho_{2}$. The ratio of the accelerations due to gravity at their surfaces is
D Given, radii of two planets is $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ Density of two planets is $\rho_{1}$ and $\rho_{2}$ For planet first $\mathrm{g}_{1}=\frac{\mathrm{Gm}_{1}}{\left(\mathrm{R}_{1}\right)^{2}}=\frac{\mathrm{G} \frac{4}{3} \pi\left(\mathrm{R}_{1}\right)^{3} \cdot \rho_{1}}{\left(\mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{1}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{1} \cdot \rho_{1}$ Similarly for second planet $\mathrm{g}_{2}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{2} \cdot \rho_{2}$ On dividing eq ${ }^{\mathrm{n}}$ (i) and (ii) we get $\frac{g_{1}}{g_{2}}=\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}$ $g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}$
MP PET 1994
Gravitation
138337
If the earth were to cease rotating about its own axis. The increase in the value of $g$ in C.G.S. system at a place of latitude of $45^{\circ}$ will be
1 2.68
2 1.68
3 3.36
4 0.34
Explanation:
B Given $\lambda=45^{\circ}, \mathrm{R}=6400 \times 10^{3} \mathrm{~m}$ $\because \quad \omega=\frac{2 \pi}{\mathrm{T}}$ $\omega=\frac{2 \pi}{24 \times 60 \times 60} \quad(\because \mathrm{T}=24 \mathrm{hr})$ The value of acceleration due to gravity with latitude $\lambda$ due to rotation of earth is, $g^{\prime}=g-R \omega^{2} \cos ^{2} \lambda$ $g^{\prime}-g=-R \omega^{2} \cos ^{2} \lambda$ $g-g^{\prime}= 6400 \times 10^{3} \times\left(\frac{2 \times 3.14}{24 \times 60 \times 60}\right)^{2} \times\left(\cos ^{2} 45^{\circ}\right)$ $g-g^{\prime}= \frac{6400 \times 10^{3} \times 4 \times 3.14 \times 3.14}{24 \times 60 \times 60 \times 24 \times 60 \times 60} \times \frac{1}{2}$ $= 16.89 \times 10^{-3} \mathrm{~m} / \mathrm{sec}^{2}$ $= 16.89 \times 10^{-1} \mathrm{~cm} / \mathrm{sec}^{2}$ $= 1.68 \mathrm{~cm} / \mathrm{sec}^{2}$
AIIMS-2010
Gravitation
138338
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is about $3200 \mathrm{~km}$. The mass of the earth is about 10 times of the mass. The object weighs $200 \mathrm{~N}$ on earth surface, then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Mass of earth $\left(M_{e}\right)=10$ time of object Weight of the object on earth $\left(\mathrm{W}_{\mathrm{e}}\right)=200 \mathrm{~N}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{Mg}_{\mathrm{m}}}{\mathrm{Mg}_{\mathrm{e}}}=\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} \times\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2} \times \mathrm{GM}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\mathrm{M}_{\mathrm{e}}\left(\mathrm{R}_{\mathrm{m}}\right)^{2}} \quad\left\{\because \mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}\right\}$ $=\frac{1}{10} \times(2)^{2} \quad\left[\because \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}=\frac{6400}{3200}=2\right]$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{4}{10}=\frac{2}{5} \quad \mathrm{~W}_{\mathrm{m}}=\mathrm{W}_{\mathrm{e}} \times \frac{2}{5}$ $\mathrm{~W}_{\mathrm{m}}=200 \times \frac{2}{5}=80 \mathrm{~N}$
AIIMS-2002
Gravitation
138335
Assertion: In a free fall, weight of a body becomes effectively zero. Reason: Acceleration due to gravity acting on a body having free fall is zero.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C In a free fall effective weight of a body becomes zero. The reason behind it, is the normal force acting on the object is equal to $\mathrm{mg}$ which is same as the weight of the object. So, the object feels weightless. So, Assertion is correct but reason is false.
138336
The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $\rho_{1}$ and $\rho_{2}$. The ratio of the accelerations due to gravity at their surfaces is
D Given, radii of two planets is $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ Density of two planets is $\rho_{1}$ and $\rho_{2}$ For planet first $\mathrm{g}_{1}=\frac{\mathrm{Gm}_{1}}{\left(\mathrm{R}_{1}\right)^{2}}=\frac{\mathrm{G} \frac{4}{3} \pi\left(\mathrm{R}_{1}\right)^{3} \cdot \rho_{1}}{\left(\mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{1}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{1} \cdot \rho_{1}$ Similarly for second planet $\mathrm{g}_{2}=\frac{4}{3} \mathrm{G} \cdot \pi \cdot \mathrm{R}_{2} \cdot \rho_{2}$ On dividing eq ${ }^{\mathrm{n}}$ (i) and (ii) we get $\frac{g_{1}}{g_{2}}=\frac{R_{1} \rho_{1}}{R_{2} \rho_{2}}$ $g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}$
MP PET 1994
Gravitation
138337
If the earth were to cease rotating about its own axis. The increase in the value of $g$ in C.G.S. system at a place of latitude of $45^{\circ}$ will be
1 2.68
2 1.68
3 3.36
4 0.34
Explanation:
B Given $\lambda=45^{\circ}, \mathrm{R}=6400 \times 10^{3} \mathrm{~m}$ $\because \quad \omega=\frac{2 \pi}{\mathrm{T}}$ $\omega=\frac{2 \pi}{24 \times 60 \times 60} \quad(\because \mathrm{T}=24 \mathrm{hr})$ The value of acceleration due to gravity with latitude $\lambda$ due to rotation of earth is, $g^{\prime}=g-R \omega^{2} \cos ^{2} \lambda$ $g^{\prime}-g=-R \omega^{2} \cos ^{2} \lambda$ $g-g^{\prime}= 6400 \times 10^{3} \times\left(\frac{2 \times 3.14}{24 \times 60 \times 60}\right)^{2} \times\left(\cos ^{2} 45^{\circ}\right)$ $g-g^{\prime}= \frac{6400 \times 10^{3} \times 4 \times 3.14 \times 3.14}{24 \times 60 \times 60 \times 24 \times 60 \times 60} \times \frac{1}{2}$ $= 16.89 \times 10^{-3} \mathrm{~m} / \mathrm{sec}^{2}$ $= 16.89 \times 10^{-1} \mathrm{~cm} / \mathrm{sec}^{2}$ $= 1.68 \mathrm{~cm} / \mathrm{sec}^{2}$
AIIMS-2010
Gravitation
138338
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is about $3200 \mathrm{~km}$. The mass of the earth is about 10 times of the mass. The object weighs $200 \mathrm{~N}$ on earth surface, then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Mass of earth $\left(M_{e}\right)=10$ time of object Weight of the object on earth $\left(\mathrm{W}_{\mathrm{e}}\right)=200 \mathrm{~N}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{Mg}_{\mathrm{m}}}{\mathrm{Mg}_{\mathrm{e}}}=\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} \times\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2} \times \mathrm{GM}_{\mathrm{e}}}$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}\left(\mathrm{R}_{\mathrm{e}}\right)^{2}}{\mathrm{M}_{\mathrm{e}}\left(\mathrm{R}_{\mathrm{m}}\right)^{2}} \quad\left\{\because \mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}\right\}$ $=\frac{1}{10} \times(2)^{2} \quad\left[\because \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}=\frac{6400}{3200}=2\right]$ $\frac{\mathrm{W}_{\mathrm{m}}}{\mathrm{W}_{\mathrm{e}}}=\frac{4}{10}=\frac{2}{5} \quad \mathrm{~W}_{\mathrm{m}}=\mathrm{W}_{\mathrm{e}} \times \frac{2}{5}$ $\mathrm{~W}_{\mathrm{m}}=200 \times \frac{2}{5}=80 \mathrm{~N}$
AIIMS-2002
Gravitation
138335
Assertion: In a free fall, weight of a body becomes effectively zero. Reason: Acceleration due to gravity acting on a body having free fall is zero.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
C In a free fall effective weight of a body becomes zero. The reason behind it, is the normal force acting on the object is equal to $\mathrm{mg}$ which is same as the weight of the object. So, the object feels weightless. So, Assertion is correct but reason is false.
