266235
In the following figure, electric flux passing through the surface is:
1 \(\frac{q}{\varepsilon_0}\)
2 \(\frac{q}{2 \varepsilon_a}\)
3 \(\frac{q}{12 \varepsilon_p}\)
4 \(\frac{q}{6 \varepsilon_\alpha}\)
Explanation:
c total flux passing through 6 surfaces. \(\phi_E=\frac{q / 2}{\varepsilon_0}\) \(\therefore\) Flux passing through one surface \(=\frac{1}{12} \frac{q}{z_0}\)
**NCERT-XII-I-29
TEST SERIES (PHYSICS FST)
266236
A stationary wave is represented by \(Y=A \sin (100 t) \cos (0.01 x)\), where \(y\) and \(A\) are in millimeters, \(t\) is in seconds, and \(x\) is in meteres, the velocity of super imposing waves is:
266237
In the figure shown below, the charge on the left plate of the \(10_\mu \mathrm{F}\) capacitor is \(-30_\mu \mathrm{C}\). The charge on the right plate of the \(6 \mu \mathrm{~F}\) capacitor is:
1 \(-18 \mu \mathrm{C}\)
2 \(-12 \mu \mathrm{C}\)
3 \(+12 \mu \mathrm{C}\)
4 \(+18 \mu \mathrm{C}\)
Explanation:
d \(6 \mu \mathrm{~F}\) and \(4 \mu \mathrm{~F}\) are in parallel and total charge on this combination is \(30 \mu \mathrm{C}\) \(\therefore\) Charge on \(6 \mu \mathrm{~F}\) capacitor \( =\frac{6}{6+4} \times 30=18 \mu \mathrm{C} \) Since charge is asked on right plate therefore is \(+18 \mu \mathrm{C}\).
**NCERT-XII-I-72**
TEST SERIES (PHYSICS FST)
266238
In stationary wave all particles between two nodes pass through the mean position:
NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
266235
In the following figure, electric flux passing through the surface is:
1 \(\frac{q}{\varepsilon_0}\)
2 \(\frac{q}{2 \varepsilon_a}\)
3 \(\frac{q}{12 \varepsilon_p}\)
4 \(\frac{q}{6 \varepsilon_\alpha}\)
Explanation:
c total flux passing through 6 surfaces. \(\phi_E=\frac{q / 2}{\varepsilon_0}\) \(\therefore\) Flux passing through one surface \(=\frac{1}{12} \frac{q}{z_0}\)
**NCERT-XII-I-29
TEST SERIES (PHYSICS FST)
266236
A stationary wave is represented by \(Y=A \sin (100 t) \cos (0.01 x)\), where \(y\) and \(A\) are in millimeters, \(t\) is in seconds, and \(x\) is in meteres, the velocity of super imposing waves is:
266237
In the figure shown below, the charge on the left plate of the \(10_\mu \mathrm{F}\) capacitor is \(-30_\mu \mathrm{C}\). The charge on the right plate of the \(6 \mu \mathrm{~F}\) capacitor is:
1 \(-18 \mu \mathrm{C}\)
2 \(-12 \mu \mathrm{C}\)
3 \(+12 \mu \mathrm{C}\)
4 \(+18 \mu \mathrm{C}\)
Explanation:
d \(6 \mu \mathrm{~F}\) and \(4 \mu \mathrm{~F}\) are in parallel and total charge on this combination is \(30 \mu \mathrm{C}\) \(\therefore\) Charge on \(6 \mu \mathrm{~F}\) capacitor \( =\frac{6}{6+4} \times 30=18 \mu \mathrm{C} \) Since charge is asked on right plate therefore is \(+18 \mu \mathrm{C}\).
**NCERT-XII-I-72**
TEST SERIES (PHYSICS FST)
266238
In stationary wave all particles between two nodes pass through the mean position:
266235
In the following figure, electric flux passing through the surface is:
1 \(\frac{q}{\varepsilon_0}\)
2 \(\frac{q}{2 \varepsilon_a}\)
3 \(\frac{q}{12 \varepsilon_p}\)
4 \(\frac{q}{6 \varepsilon_\alpha}\)
Explanation:
c total flux passing through 6 surfaces. \(\phi_E=\frac{q / 2}{\varepsilon_0}\) \(\therefore\) Flux passing through one surface \(=\frac{1}{12} \frac{q}{z_0}\)
**NCERT-XII-I-29
TEST SERIES (PHYSICS FST)
266236
A stationary wave is represented by \(Y=A \sin (100 t) \cos (0.01 x)\), where \(y\) and \(A\) are in millimeters, \(t\) is in seconds, and \(x\) is in meteres, the velocity of super imposing waves is:
266237
In the figure shown below, the charge on the left plate of the \(10_\mu \mathrm{F}\) capacitor is \(-30_\mu \mathrm{C}\). The charge on the right plate of the \(6 \mu \mathrm{~F}\) capacitor is:
1 \(-18 \mu \mathrm{C}\)
2 \(-12 \mu \mathrm{C}\)
3 \(+12 \mu \mathrm{C}\)
4 \(+18 \mu \mathrm{C}\)
Explanation:
d \(6 \mu \mathrm{~F}\) and \(4 \mu \mathrm{~F}\) are in parallel and total charge on this combination is \(30 \mu \mathrm{C}\) \(\therefore\) Charge on \(6 \mu \mathrm{~F}\) capacitor \( =\frac{6}{6+4} \times 30=18 \mu \mathrm{C} \) Since charge is asked on right plate therefore is \(+18 \mu \mathrm{C}\).
**NCERT-XII-I-72**
TEST SERIES (PHYSICS FST)
266238
In stationary wave all particles between two nodes pass through the mean position:
266235
In the following figure, electric flux passing through the surface is:
1 \(\frac{q}{\varepsilon_0}\)
2 \(\frac{q}{2 \varepsilon_a}\)
3 \(\frac{q}{12 \varepsilon_p}\)
4 \(\frac{q}{6 \varepsilon_\alpha}\)
Explanation:
c total flux passing through 6 surfaces. \(\phi_E=\frac{q / 2}{\varepsilon_0}\) \(\therefore\) Flux passing through one surface \(=\frac{1}{12} \frac{q}{z_0}\)
**NCERT-XII-I-29
TEST SERIES (PHYSICS FST)
266236
A stationary wave is represented by \(Y=A \sin (100 t) \cos (0.01 x)\), where \(y\) and \(A\) are in millimeters, \(t\) is in seconds, and \(x\) is in meteres, the velocity of super imposing waves is:
266237
In the figure shown below, the charge on the left plate of the \(10_\mu \mathrm{F}\) capacitor is \(-30_\mu \mathrm{C}\). The charge on the right plate of the \(6 \mu \mathrm{~F}\) capacitor is:
1 \(-18 \mu \mathrm{C}\)
2 \(-12 \mu \mathrm{C}\)
3 \(+12 \mu \mathrm{C}\)
4 \(+18 \mu \mathrm{C}\)
Explanation:
d \(6 \mu \mathrm{~F}\) and \(4 \mu \mathrm{~F}\) are in parallel and total charge on this combination is \(30 \mu \mathrm{C}\) \(\therefore\) Charge on \(6 \mu \mathrm{~F}\) capacitor \( =\frac{6}{6+4} \times 30=18 \mu \mathrm{C} \) Since charge is asked on right plate therefore is \(+18 \mu \mathrm{C}\).
**NCERT-XII-I-72**
TEST SERIES (PHYSICS FST)
266238
In stationary wave all particles between two nodes pass through the mean position:
