263827
The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:
1 \(\frac{1}{2} \mathrm{CV}^2\)
2 \(\mathrm{Cl}^2\)
3 \(\frac{2}{3} \mathrm{CV}^2\)
4 \(\frac{3}{2} \mathrm{CV}^2\)
Explanation:
d Net capacity \(\mathrm{C}_{\text {Het }}=3 \mathrm{C}\) \[ \begin{aligned} & U=\frac{1}{2} c_{\text {H }} V^2 \\ & U=\frac{1}{2}(3 C) y^2 \quad \Rightarrow U=\frac{3}{2} C V^2 \end{aligned} \]
NCERT-XII-I-73
TEST SERIES (PHYSICS FST)
263828
A satellite revolve very near to the earth (radius \(\mathrm{R}_{\mathrm{e}}\) ) then it's time - period is:
1 \(\pi \sqrt{\frac{2 \mathrm{R}_e}{g}}\)
2 \(2 \pi \sqrt{\frac{2 R_e}{g}}\)
3 \(2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
4 \(\frac{\pi}{2} \sqrt{\frac{\mathrm{R}_e}{g}}\)
Explanation:
c The time period at height h from earth's surface is . \[ \mathrm{T}=2 \pi \sqrt{\frac{\left(\mathrm{R}_e+h\right)^3}{g \mathrm{R}_e^2}} \] for very пear \(h \rightarrow 0\) so \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
NCERT-XI-I-137
TEST SERIES (PHYSICS FST)
263829
The sinusoidalA.C. current flows through a register of resistance \(R\). If the peak current is \(I_p\). then power dissipated is:
263827
The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:
1 \(\frac{1}{2} \mathrm{CV}^2\)
2 \(\mathrm{Cl}^2\)
3 \(\frac{2}{3} \mathrm{CV}^2\)
4 \(\frac{3}{2} \mathrm{CV}^2\)
Explanation:
d Net capacity \(\mathrm{C}_{\text {Het }}=3 \mathrm{C}\) \[ \begin{aligned} & U=\frac{1}{2} c_{\text {H }} V^2 \\ & U=\frac{1}{2}(3 C) y^2 \quad \Rightarrow U=\frac{3}{2} C V^2 \end{aligned} \]
NCERT-XII-I-73
TEST SERIES (PHYSICS FST)
263828
A satellite revolve very near to the earth (radius \(\mathrm{R}_{\mathrm{e}}\) ) then it's time - period is:
1 \(\pi \sqrt{\frac{2 \mathrm{R}_e}{g}}\)
2 \(2 \pi \sqrt{\frac{2 R_e}{g}}\)
3 \(2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
4 \(\frac{\pi}{2} \sqrt{\frac{\mathrm{R}_e}{g}}\)
Explanation:
c The time period at height h from earth's surface is . \[ \mathrm{T}=2 \pi \sqrt{\frac{\left(\mathrm{R}_e+h\right)^3}{g \mathrm{R}_e^2}} \] for very пear \(h \rightarrow 0\) so \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
NCERT-XI-I-137
TEST SERIES (PHYSICS FST)
263829
The sinusoidalA.C. current flows through a register of resistance \(R\). If the peak current is \(I_p\). then power dissipated is:
263827
The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:
1 \(\frac{1}{2} \mathrm{CV}^2\)
2 \(\mathrm{Cl}^2\)
3 \(\frac{2}{3} \mathrm{CV}^2\)
4 \(\frac{3}{2} \mathrm{CV}^2\)
Explanation:
d Net capacity \(\mathrm{C}_{\text {Het }}=3 \mathrm{C}\) \[ \begin{aligned} & U=\frac{1}{2} c_{\text {H }} V^2 \\ & U=\frac{1}{2}(3 C) y^2 \quad \Rightarrow U=\frac{3}{2} C V^2 \end{aligned} \]
NCERT-XII-I-73
TEST SERIES (PHYSICS FST)
263828
A satellite revolve very near to the earth (radius \(\mathrm{R}_{\mathrm{e}}\) ) then it's time - period is:
1 \(\pi \sqrt{\frac{2 \mathrm{R}_e}{g}}\)
2 \(2 \pi \sqrt{\frac{2 R_e}{g}}\)
3 \(2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
4 \(\frac{\pi}{2} \sqrt{\frac{\mathrm{R}_e}{g}}\)
Explanation:
c The time period at height h from earth's surface is . \[ \mathrm{T}=2 \pi \sqrt{\frac{\left(\mathrm{R}_e+h\right)^3}{g \mathrm{R}_e^2}} \] for very пear \(h \rightarrow 0\) so \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
NCERT-XI-I-137
TEST SERIES (PHYSICS FST)
263829
The sinusoidalA.C. current flows through a register of resistance \(R\). If the peak current is \(I_p\). then power dissipated is:
263827
The capacitors are uncharged in the begining . How much heat will be produced, when switch S is closed:
1 \(\frac{1}{2} \mathrm{CV}^2\)
2 \(\mathrm{Cl}^2\)
3 \(\frac{2}{3} \mathrm{CV}^2\)
4 \(\frac{3}{2} \mathrm{CV}^2\)
Explanation:
d Net capacity \(\mathrm{C}_{\text {Het }}=3 \mathrm{C}\) \[ \begin{aligned} & U=\frac{1}{2} c_{\text {H }} V^2 \\ & U=\frac{1}{2}(3 C) y^2 \quad \Rightarrow U=\frac{3}{2} C V^2 \end{aligned} \]
NCERT-XII-I-73
TEST SERIES (PHYSICS FST)
263828
A satellite revolve very near to the earth (radius \(\mathrm{R}_{\mathrm{e}}\) ) then it's time - period is:
1 \(\pi \sqrt{\frac{2 \mathrm{R}_e}{g}}\)
2 \(2 \pi \sqrt{\frac{2 R_e}{g}}\)
3 \(2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
4 \(\frac{\pi}{2} \sqrt{\frac{\mathrm{R}_e}{g}}\)
Explanation:
c The time period at height h from earth's surface is . \[ \mathrm{T}=2 \pi \sqrt{\frac{\left(\mathrm{R}_e+h\right)^3}{g \mathrm{R}_e^2}} \] for very пear \(h \rightarrow 0\) so \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}_e}{g}}\)
NCERT-XI-I-137
TEST SERIES (PHYSICS FST)
263829
The sinusoidalA.C. current flows through a register of resistance \(R\). If the peak current is \(I_p\). then power dissipated is:
