263814
The v-t graph for a particle is shown. The distance travelled in the first four seconds is:
1 12 m
2 16 m
3 20 m
4 24 m
Explanation:
b distance travelled = Area under \(v-\mathrm{t}\) graph \[ =\frac{1}{2} \times 2 \times 8+\frac{1}{2} \times 2 \times 8=16 \]
NCERT-XI-I-16
TEST SERIES (PHYSICS FST)
263815
A polarized light of intensity \(\mathbf{I}_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emerging polarised light from second polarizer?
263816
In shown circuit what should be the value of resistance that should be attached in parallel with 20 \$2 so that no current flows through 59 resistance:
1 \(\frac{40}{3} \Omega\)
2 \(\frac{80}{3} \Omega\)
3 \(\frac{100}{3} \Omega\)
4 \(8 \Omega\)
Explanation:
a Let us apply condition of balanced wheat stone brige. \(\frac{4}{\phi}=\frac{6}{12} \Rightarrow \frac{4}{\phi}=\frac{1}{2} \Rightarrow \phi=8 \Omega\) soQR must have \(8 \Omega\) now let \(x_{\Omega}\) resistance must be attacted in parallel so. \[ \begin{gathered} \frac{1}{8}=\frac{1}{x}+\frac{1}{20} \Rightarrow \frac{1}{x}=\frac{1}{8}-\frac{1}{20}=\frac{5-2}{40}=\frac{3}{40} \\ x=\frac{40}{3} \Omega \end{gathered} \]
NCERT-XII-I-100
TEST SERIES (PHYSICS FST)
263817
In equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\). The quantity \(\frac{a}{P V b}\) is dimensionally equal to :
NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
263814
The v-t graph for a particle is shown. The distance travelled in the first four seconds is:
1 12 m
2 16 m
3 20 m
4 24 m
Explanation:
b distance travelled = Area under \(v-\mathrm{t}\) graph \[ =\frac{1}{2} \times 2 \times 8+\frac{1}{2} \times 2 \times 8=16 \]
NCERT-XI-I-16
TEST SERIES (PHYSICS FST)
263815
A polarized light of intensity \(\mathbf{I}_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emerging polarised light from second polarizer?
263816
In shown circuit what should be the value of resistance that should be attached in parallel with 20 \$2 so that no current flows through 59 resistance:
1 \(\frac{40}{3} \Omega\)
2 \(\frac{80}{3} \Omega\)
3 \(\frac{100}{3} \Omega\)
4 \(8 \Omega\)
Explanation:
a Let us apply condition of balanced wheat stone brige. \(\frac{4}{\phi}=\frac{6}{12} \Rightarrow \frac{4}{\phi}=\frac{1}{2} \Rightarrow \phi=8 \Omega\) soQR must have \(8 \Omega\) now let \(x_{\Omega}\) resistance must be attacted in parallel so. \[ \begin{gathered} \frac{1}{8}=\frac{1}{x}+\frac{1}{20} \Rightarrow \frac{1}{x}=\frac{1}{8}-\frac{1}{20}=\frac{5-2}{40}=\frac{3}{40} \\ x=\frac{40}{3} \Omega \end{gathered} \]
NCERT-XII-I-100
TEST SERIES (PHYSICS FST)
263817
In equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\). The quantity \(\frac{a}{P V b}\) is dimensionally equal to :
263814
The v-t graph for a particle is shown. The distance travelled in the first four seconds is:
1 12 m
2 16 m
3 20 m
4 24 m
Explanation:
b distance travelled = Area under \(v-\mathrm{t}\) graph \[ =\frac{1}{2} \times 2 \times 8+\frac{1}{2} \times 2 \times 8=16 \]
NCERT-XI-I-16
TEST SERIES (PHYSICS FST)
263815
A polarized light of intensity \(\mathbf{I}_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emerging polarised light from second polarizer?
263816
In shown circuit what should be the value of resistance that should be attached in parallel with 20 \$2 so that no current flows through 59 resistance:
1 \(\frac{40}{3} \Omega\)
2 \(\frac{80}{3} \Omega\)
3 \(\frac{100}{3} \Omega\)
4 \(8 \Omega\)
Explanation:
a Let us apply condition of balanced wheat stone brige. \(\frac{4}{\phi}=\frac{6}{12} \Rightarrow \frac{4}{\phi}=\frac{1}{2} \Rightarrow \phi=8 \Omega\) soQR must have \(8 \Omega\) now let \(x_{\Omega}\) resistance must be attacted in parallel so. \[ \begin{gathered} \frac{1}{8}=\frac{1}{x}+\frac{1}{20} \Rightarrow \frac{1}{x}=\frac{1}{8}-\frac{1}{20}=\frac{5-2}{40}=\frac{3}{40} \\ x=\frac{40}{3} \Omega \end{gathered} \]
NCERT-XII-I-100
TEST SERIES (PHYSICS FST)
263817
In equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\). The quantity \(\frac{a}{P V b}\) is dimensionally equal to :
263814
The v-t graph for a particle is shown. The distance travelled in the first four seconds is:
1 12 m
2 16 m
3 20 m
4 24 m
Explanation:
b distance travelled = Area under \(v-\mathrm{t}\) graph \[ =\frac{1}{2} \times 2 \times 8+\frac{1}{2} \times 2 \times 8=16 \]
NCERT-XI-I-16
TEST SERIES (PHYSICS FST)
263815
A polarized light of intensity \(\mathbf{I}_0\) is passed through another polarizer whose pass axis makes an angle of \(60^{\circ}\) with the pass axis of the former. What is the intensity of emerging polarised light from second polarizer?
263816
In shown circuit what should be the value of resistance that should be attached in parallel with 20 \$2 so that no current flows through 59 resistance:
1 \(\frac{40}{3} \Omega\)
2 \(\frac{80}{3} \Omega\)
3 \(\frac{100}{3} \Omega\)
4 \(8 \Omega\)
Explanation:
a Let us apply condition of balanced wheat stone brige. \(\frac{4}{\phi}=\frac{6}{12} \Rightarrow \frac{4}{\phi}=\frac{1}{2} \Rightarrow \phi=8 \Omega\) soQR must have \(8 \Omega\) now let \(x_{\Omega}\) resistance must be attacted in parallel so. \[ \begin{gathered} \frac{1}{8}=\frac{1}{x}+\frac{1}{20} \Rightarrow \frac{1}{x}=\frac{1}{8}-\frac{1}{20}=\frac{5-2}{40}=\frac{3}{40} \\ x=\frac{40}{3} \Omega \end{gathered} \]
NCERT-XII-I-100
TEST SERIES (PHYSICS FST)
263817
In equation \(\left(P+\frac{a}{V^2}\right)(V-b)=R T\). The quantity \(\frac{a}{P V b}\) is dimensionally equal to :
