NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
266526
The thermal capacity of a body is 120 cal , then its water equivalent is :
1 120 call g
2 120 g
3 12 g
4 12 kg .
Explanation:
b We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body.
**NCERT - XI-II-208**
TEST SERIES (PHYSICS FST)
266527
Assertion - Radiation is the speediest mode of heat transfer. Reason - Radiation is transmitted in Zig-Zag motion.
1 Assertion (A) is true \& reason (R) is false.
2 \(A\) is true \& \(B\) is also true.
3 \(A\) is false \& \(B\) is true
4 None of these
Explanation:
a Thermal Radiation is always transmitted in a straight line.
**NCERT - XI-II-218**
TEST SERIES (PHYSICS FST)
266528
The Radius of the earth is 6400 km and \(\mathrm{g}=10 \mathrm{ml}\) \(\mathrm{sec}^2\) in order that a body of 5 kg weight zero at the equator, the angular speed of the earth is-
266529
A geo-stationary satellite is orbiting the earth at a height of 8 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 3.5 R from the surface of earth is:
1 10 hr
2 \((6 / \sqrt{2}) \mathrm{hr}\)
3 6 hr
4 \(6 \sqrt{2} \mathrm{hr}\)
Explanation:
d Distance of planet from the centre ofearth is \(9 R\) and 4.5R \[ \begin{aligned} & \mathrm{T}^2 \propto \mathrm{R}^3 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{4.5 \mathrm{R}}{9 \mathrm{R}}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{1}{2}\right)^{\frac{3}{2}} \\ & \mathrm{~T}_2=24 \times \frac{1}{2 \sqrt{2}} \\ & \mathrm{~T}_2=\frac{12}{\sqrt{2}}=6 \sqrt{2 \mathrm{H}} \mathrm{r} \end{aligned} \]
266526
The thermal capacity of a body is 120 cal , then its water equivalent is :
1 120 call g
2 120 g
3 12 g
4 12 kg .
Explanation:
b We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body.
**NCERT - XI-II-208**
TEST SERIES (PHYSICS FST)
266527
Assertion - Radiation is the speediest mode of heat transfer. Reason - Radiation is transmitted in Zig-Zag motion.
1 Assertion (A) is true \& reason (R) is false.
2 \(A\) is true \& \(B\) is also true.
3 \(A\) is false \& \(B\) is true
4 None of these
Explanation:
a Thermal Radiation is always transmitted in a straight line.
**NCERT - XI-II-218**
TEST SERIES (PHYSICS FST)
266528
The Radius of the earth is 6400 km and \(\mathrm{g}=10 \mathrm{ml}\) \(\mathrm{sec}^2\) in order that a body of 5 kg weight zero at the equator, the angular speed of the earth is-
266529
A geo-stationary satellite is orbiting the earth at a height of 8 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 3.5 R from the surface of earth is:
1 10 hr
2 \((6 / \sqrt{2}) \mathrm{hr}\)
3 6 hr
4 \(6 \sqrt{2} \mathrm{hr}\)
Explanation:
d Distance of planet from the centre ofearth is \(9 R\) and 4.5R \[ \begin{aligned} & \mathrm{T}^2 \propto \mathrm{R}^3 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{4.5 \mathrm{R}}{9 \mathrm{R}}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{1}{2}\right)^{\frac{3}{2}} \\ & \mathrm{~T}_2=24 \times \frac{1}{2 \sqrt{2}} \\ & \mathrm{~T}_2=\frac{12}{\sqrt{2}}=6 \sqrt{2 \mathrm{H}} \mathrm{r} \end{aligned} \]
266526
The thermal capacity of a body is 120 cal , then its water equivalent is :
1 120 call g
2 120 g
3 12 g
4 12 kg .
Explanation:
b We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body.
**NCERT - XI-II-208**
TEST SERIES (PHYSICS FST)
266527
Assertion - Radiation is the speediest mode of heat transfer. Reason - Radiation is transmitted in Zig-Zag motion.
1 Assertion (A) is true \& reason (R) is false.
2 \(A\) is true \& \(B\) is also true.
3 \(A\) is false \& \(B\) is true
4 None of these
Explanation:
a Thermal Radiation is always transmitted in a straight line.
**NCERT - XI-II-218**
TEST SERIES (PHYSICS FST)
266528
The Radius of the earth is 6400 km and \(\mathrm{g}=10 \mathrm{ml}\) \(\mathrm{sec}^2\) in order that a body of 5 kg weight zero at the equator, the angular speed of the earth is-
266529
A geo-stationary satellite is orbiting the earth at a height of 8 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 3.5 R from the surface of earth is:
1 10 hr
2 \((6 / \sqrt{2}) \mathrm{hr}\)
3 6 hr
4 \(6 \sqrt{2} \mathrm{hr}\)
Explanation:
d Distance of planet from the centre ofearth is \(9 R\) and 4.5R \[ \begin{aligned} & \mathrm{T}^2 \propto \mathrm{R}^3 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{4.5 \mathrm{R}}{9 \mathrm{R}}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{1}{2}\right)^{\frac{3}{2}} \\ & \mathrm{~T}_2=24 \times \frac{1}{2 \sqrt{2}} \\ & \mathrm{~T}_2=\frac{12}{\sqrt{2}}=6 \sqrt{2 \mathrm{H}} \mathrm{r} \end{aligned} \]
266526
The thermal capacity of a body is 120 cal , then its water equivalent is :
1 120 call g
2 120 g
3 12 g
4 12 kg .
Explanation:
b We know that thermal capacity of a body expressed in calories is equal to water equivalent of the body.
**NCERT - XI-II-208**
TEST SERIES (PHYSICS FST)
266527
Assertion - Radiation is the speediest mode of heat transfer. Reason - Radiation is transmitted in Zig-Zag motion.
1 Assertion (A) is true \& reason (R) is false.
2 \(A\) is true \& \(B\) is also true.
3 \(A\) is false \& \(B\) is true
4 None of these
Explanation:
a Thermal Radiation is always transmitted in a straight line.
**NCERT - XI-II-218**
TEST SERIES (PHYSICS FST)
266528
The Radius of the earth is 6400 km and \(\mathrm{g}=10 \mathrm{ml}\) \(\mathrm{sec}^2\) in order that a body of 5 kg weight zero at the equator, the angular speed of the earth is-
266529
A geo-stationary satellite is orbiting the earth at a height of 8 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 3.5 R from the surface of earth is:
1 10 hr
2 \((6 / \sqrt{2}) \mathrm{hr}\)
3 6 hr
4 \(6 \sqrt{2} \mathrm{hr}\)
Explanation:
d Distance of planet from the centre ofearth is \(9 R\) and 4.5R \[ \begin{aligned} & \mathrm{T}^2 \propto \mathrm{R}^3 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{4.5 \mathrm{R}}{9 \mathrm{R}}\right)^{3 / 2} \\ & \mathrm{~T}_2=24 \times\left(\frac{1}{2}\right)^{\frac{3}{2}} \\ & \mathrm{~T}_2=24 \times \frac{1}{2 \sqrt{2}} \\ & \mathrm{~T}_2=\frac{12}{\sqrt{2}}=6 \sqrt{2 \mathrm{H}} \mathrm{r} \end{aligned} \]
