266272
A particle in a certain conservative force field has a. potential energy given by \(\mathbf{U}=\frac{1}{2}\left(x^2-z^2\right)\). The force exerted on it is:
1 \(-x \hat{i}+z \hat{k}\)
2 \(x \hat{i}+z \hat{k}\)
3 \(x \hat{i}-z \hat{k}\)
4 \(\frac{1}{2}(x \hat{i}-z \hat{k})\)
Explanation:
a Given: \(U=\frac{1}{2}\left(x^2-z^2\right)\) For a conservative field \(F=-\frac{d U}{d x}\) \[ \begin{aligned} & \mathrm{F}_{\mathrm{x}}=\frac{-\partial U}{\partial \mathrm{x}}=\frac{-\partial}{\partial x}\left[\frac{\left(\mathrm{x}^2-z^2\right)}{2}\right]=-\mathrm{x} \\ & \mathrm{~F}_{\mathrm{z}}=\frac{-\partial U}{\partial z}=\frac{-\partial}{\partial z}\left[\frac{\left(x^2-z^2\right)}{2}\right]=+z \end{aligned} \] \[ F=F_x+F_2 \]
**NCERT-XI-I-78**
TEST SERIES (PHYSICS FST)
266273
Star \(A\) has radius r surface temperature \(T\) while star B has radius \(4 r\) and surface temperature \(\mathrm{T} / 2\). The ratio of the radiation power emitted by stars, \(P_{\mathrm{\alpha}}: \mathrm{P}_{\mathrm{e}}\) is:
1 \(16: 1\)
2 \(1: 16\)
3 \(1: 1\)
4 \(1: 4\)
Explanation:
c Fadiation Power \(P\) a \(\Gamma^2 T^4\) \[ \frac{P_B}{P_A}=\left(\frac{r_B}{r_A}\right)^2 \times\left(\frac{T_B}{T_A}\right)^4=\left(\frac{4 r}{r}\right)^2 \times\left(\frac{T / 2}{T}\right)^4=1 \]
**NCBRT-XI-II-219**
TEST SERIES (PHYSICS FST)
266274
The force of attraction between masses of two bodies is 100 N . If mass of one body is increased by \(10 \%\) and mass of another body decreased by \(10 \%\) without changing distance between them, then new force of attraction will be:
1 100 N
2 99 N
3 90 N
4 None of these
Explanation:
b \(F=\frac{G m_1 m_2}{r^2}=100 \mathrm{~N} \) If mass of one body is increased by \(10 \%\) and mass of another body is decreased by \(10 \%\) \[ \begin{aligned} & m_1^{\prime}=m_1+10 \% \text { of } m_1=m_1+0.1 m_1=1.1 m_1 \\ & m_2^{\prime}=m_2-10 \% \text { of } m_2=m_2-0.1 m_2=0.9 m_2 \\ & F^{\prime}=\frac{G m_1^{\prime} m_2^{\prime}}{r^2}=\frac{G\left(1.1 m_1\right)\left(0.9 m_2\right)}{r^2} \\ & =\frac{G m_1 m_2}{r^2} \times 1.1 \times 0.9=0.99 \times 100=99 \mathrm{~N} \end{aligned} \]
**NCERT-XI-I-130**
TEST SERIES (PHYSICS FST)
266275
In half-wave rectifier, the input frequency is 50 Hz . What is the output frequency of a full wave rectifier for the same input frequency.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
TEST SERIES (PHYSICS FST)
266272
A particle in a certain conservative force field has a. potential energy given by \(\mathbf{U}=\frac{1}{2}\left(x^2-z^2\right)\). The force exerted on it is:
1 \(-x \hat{i}+z \hat{k}\)
2 \(x \hat{i}+z \hat{k}\)
3 \(x \hat{i}-z \hat{k}\)
4 \(\frac{1}{2}(x \hat{i}-z \hat{k})\)
Explanation:
a Given: \(U=\frac{1}{2}\left(x^2-z^2\right)\) For a conservative field \(F=-\frac{d U}{d x}\) \[ \begin{aligned} & \mathrm{F}_{\mathrm{x}}=\frac{-\partial U}{\partial \mathrm{x}}=\frac{-\partial}{\partial x}\left[\frac{\left(\mathrm{x}^2-z^2\right)}{2}\right]=-\mathrm{x} \\ & \mathrm{~F}_{\mathrm{z}}=\frac{-\partial U}{\partial z}=\frac{-\partial}{\partial z}\left[\frac{\left(x^2-z^2\right)}{2}\right]=+z \end{aligned} \] \[ F=F_x+F_2 \]
**NCERT-XI-I-78**
TEST SERIES (PHYSICS FST)
266273
Star \(A\) has radius r surface temperature \(T\) while star B has radius \(4 r\) and surface temperature \(\mathrm{T} / 2\). The ratio of the radiation power emitted by stars, \(P_{\mathrm{\alpha}}: \mathrm{P}_{\mathrm{e}}\) is:
1 \(16: 1\)
2 \(1: 16\)
3 \(1: 1\)
4 \(1: 4\)
Explanation:
c Fadiation Power \(P\) a \(\Gamma^2 T^4\) \[ \frac{P_B}{P_A}=\left(\frac{r_B}{r_A}\right)^2 \times\left(\frac{T_B}{T_A}\right)^4=\left(\frac{4 r}{r}\right)^2 \times\left(\frac{T / 2}{T}\right)^4=1 \]
**NCBRT-XI-II-219**
TEST SERIES (PHYSICS FST)
266274
The force of attraction between masses of two bodies is 100 N . If mass of one body is increased by \(10 \%\) and mass of another body decreased by \(10 \%\) without changing distance between them, then new force of attraction will be:
1 100 N
2 99 N
3 90 N
4 None of these
Explanation:
b \(F=\frac{G m_1 m_2}{r^2}=100 \mathrm{~N} \) If mass of one body is increased by \(10 \%\) and mass of another body is decreased by \(10 \%\) \[ \begin{aligned} & m_1^{\prime}=m_1+10 \% \text { of } m_1=m_1+0.1 m_1=1.1 m_1 \\ & m_2^{\prime}=m_2-10 \% \text { of } m_2=m_2-0.1 m_2=0.9 m_2 \\ & F^{\prime}=\frac{G m_1^{\prime} m_2^{\prime}}{r^2}=\frac{G\left(1.1 m_1\right)\left(0.9 m_2\right)}{r^2} \\ & =\frac{G m_1 m_2}{r^2} \times 1.1 \times 0.9=0.99 \times 100=99 \mathrm{~N} \end{aligned} \]
**NCERT-XI-I-130**
TEST SERIES (PHYSICS FST)
266275
In half-wave rectifier, the input frequency is 50 Hz . What is the output frequency of a full wave rectifier for the same input frequency.
266272
A particle in a certain conservative force field has a. potential energy given by \(\mathbf{U}=\frac{1}{2}\left(x^2-z^2\right)\). The force exerted on it is:
1 \(-x \hat{i}+z \hat{k}\)
2 \(x \hat{i}+z \hat{k}\)
3 \(x \hat{i}-z \hat{k}\)
4 \(\frac{1}{2}(x \hat{i}-z \hat{k})\)
Explanation:
a Given: \(U=\frac{1}{2}\left(x^2-z^2\right)\) For a conservative field \(F=-\frac{d U}{d x}\) \[ \begin{aligned} & \mathrm{F}_{\mathrm{x}}=\frac{-\partial U}{\partial \mathrm{x}}=\frac{-\partial}{\partial x}\left[\frac{\left(\mathrm{x}^2-z^2\right)}{2}\right]=-\mathrm{x} \\ & \mathrm{~F}_{\mathrm{z}}=\frac{-\partial U}{\partial z}=\frac{-\partial}{\partial z}\left[\frac{\left(x^2-z^2\right)}{2}\right]=+z \end{aligned} \] \[ F=F_x+F_2 \]
**NCERT-XI-I-78**
TEST SERIES (PHYSICS FST)
266273
Star \(A\) has radius r surface temperature \(T\) while star B has radius \(4 r\) and surface temperature \(\mathrm{T} / 2\). The ratio of the radiation power emitted by stars, \(P_{\mathrm{\alpha}}: \mathrm{P}_{\mathrm{e}}\) is:
1 \(16: 1\)
2 \(1: 16\)
3 \(1: 1\)
4 \(1: 4\)
Explanation:
c Fadiation Power \(P\) a \(\Gamma^2 T^4\) \[ \frac{P_B}{P_A}=\left(\frac{r_B}{r_A}\right)^2 \times\left(\frac{T_B}{T_A}\right)^4=\left(\frac{4 r}{r}\right)^2 \times\left(\frac{T / 2}{T}\right)^4=1 \]
**NCBRT-XI-II-219**
TEST SERIES (PHYSICS FST)
266274
The force of attraction between masses of two bodies is 100 N . If mass of one body is increased by \(10 \%\) and mass of another body decreased by \(10 \%\) without changing distance between them, then new force of attraction will be:
1 100 N
2 99 N
3 90 N
4 None of these
Explanation:
b \(F=\frac{G m_1 m_2}{r^2}=100 \mathrm{~N} \) If mass of one body is increased by \(10 \%\) and mass of another body is decreased by \(10 \%\) \[ \begin{aligned} & m_1^{\prime}=m_1+10 \% \text { of } m_1=m_1+0.1 m_1=1.1 m_1 \\ & m_2^{\prime}=m_2-10 \% \text { of } m_2=m_2-0.1 m_2=0.9 m_2 \\ & F^{\prime}=\frac{G m_1^{\prime} m_2^{\prime}}{r^2}=\frac{G\left(1.1 m_1\right)\left(0.9 m_2\right)}{r^2} \\ & =\frac{G m_1 m_2}{r^2} \times 1.1 \times 0.9=0.99 \times 100=99 \mathrm{~N} \end{aligned} \]
**NCERT-XI-I-130**
TEST SERIES (PHYSICS FST)
266275
In half-wave rectifier, the input frequency is 50 Hz . What is the output frequency of a full wave rectifier for the same input frequency.
266272
A particle in a certain conservative force field has a. potential energy given by \(\mathbf{U}=\frac{1}{2}\left(x^2-z^2\right)\). The force exerted on it is:
1 \(-x \hat{i}+z \hat{k}\)
2 \(x \hat{i}+z \hat{k}\)
3 \(x \hat{i}-z \hat{k}\)
4 \(\frac{1}{2}(x \hat{i}-z \hat{k})\)
Explanation:
a Given: \(U=\frac{1}{2}\left(x^2-z^2\right)\) For a conservative field \(F=-\frac{d U}{d x}\) \[ \begin{aligned} & \mathrm{F}_{\mathrm{x}}=\frac{-\partial U}{\partial \mathrm{x}}=\frac{-\partial}{\partial x}\left[\frac{\left(\mathrm{x}^2-z^2\right)}{2}\right]=-\mathrm{x} \\ & \mathrm{~F}_{\mathrm{z}}=\frac{-\partial U}{\partial z}=\frac{-\partial}{\partial z}\left[\frac{\left(x^2-z^2\right)}{2}\right]=+z \end{aligned} \] \[ F=F_x+F_2 \]
**NCERT-XI-I-78**
TEST SERIES (PHYSICS FST)
266273
Star \(A\) has radius r surface temperature \(T\) while star B has radius \(4 r\) and surface temperature \(\mathrm{T} / 2\). The ratio of the radiation power emitted by stars, \(P_{\mathrm{\alpha}}: \mathrm{P}_{\mathrm{e}}\) is:
1 \(16: 1\)
2 \(1: 16\)
3 \(1: 1\)
4 \(1: 4\)
Explanation:
c Fadiation Power \(P\) a \(\Gamma^2 T^4\) \[ \frac{P_B}{P_A}=\left(\frac{r_B}{r_A}\right)^2 \times\left(\frac{T_B}{T_A}\right)^4=\left(\frac{4 r}{r}\right)^2 \times\left(\frac{T / 2}{T}\right)^4=1 \]
**NCBRT-XI-II-219**
TEST SERIES (PHYSICS FST)
266274
The force of attraction between masses of two bodies is 100 N . If mass of one body is increased by \(10 \%\) and mass of another body decreased by \(10 \%\) without changing distance between them, then new force of attraction will be:
1 100 N
2 99 N
3 90 N
4 None of these
Explanation:
b \(F=\frac{G m_1 m_2}{r^2}=100 \mathrm{~N} \) If mass of one body is increased by \(10 \%\) and mass of another body is decreased by \(10 \%\) \[ \begin{aligned} & m_1^{\prime}=m_1+10 \% \text { of } m_1=m_1+0.1 m_1=1.1 m_1 \\ & m_2^{\prime}=m_2-10 \% \text { of } m_2=m_2-0.1 m_2=0.9 m_2 \\ & F^{\prime}=\frac{G m_1^{\prime} m_2^{\prime}}{r^2}=\frac{G\left(1.1 m_1\right)\left(0.9 m_2\right)}{r^2} \\ & =\frac{G m_1 m_2}{r^2} \times 1.1 \times 0.9=0.99 \times 100=99 \mathrm{~N} \end{aligned} \]
**NCERT-XI-I-130**
TEST SERIES (PHYSICS FST)
266275
In half-wave rectifier, the input frequency is 50 Hz . What is the output frequency of a full wave rectifier for the same input frequency.