266579
If graph between binding energy per nuclion with mass no. \(A\) is shown in figure there are 4 elements P, Q, R, S, shown on the graph then select the correct option :
1 P, Q may produce fussion reaction
2 P,Q may produce fission reaction
3 R,S, may produce fission reaction
4 Both \(A \& C\)
Explanation:
d \(\mathrm{P}, \mathrm{Q}\) nuclear fussion reaction shows R S nuclear fission reaction shows
**NCERT-450**
TEST SERIES (PHYSICS FST)
266580
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result or photon emission will be:
c According to law of conservation of momentum Momentum of photon \(=\) Momentum of atom or \[ \frac{h}{\lambda}=m v \]
**NCERT-427**
TEST SERIES (PHYSICS FST)
266581
In the figure (i) \& (ii) two vibrating air-colums of same length are shown. The ratio of frequencies in the two cases will be:
1 \(1: 2\)
2 \(3: 5\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
b \[ \begin{aligned} v=\frac{h}{\lambda m} & =\frac{h}{m} z^2 R\left[\frac{1}{t^2}-\frac{1}{5^2}\right] \\ & =\frac{h}{m} R\left[1-\frac{1}{25}\right]=\frac{24}{25} \frac{h R}{m} \end{aligned} \] In closed pipe. The frequencies are in odd ratio so \[ n_1: n_2: n_3: \ldots \ldots=1: 3: 5 \] Hence in case (i) \& (ii) the frequency ratio are 3 : 5
**NCERT -374**
TEST SERIES (PHYSICS FST)
266582
Four charges equal to \(-Q\) are placed at the four corners of a square and a charge q is placed at its centre. If the system is in equilibrium, the value of \(q\) is:
1 \(-\frac{Q}{4}(1+2 \sqrt{2})\)
2 \(\frac{Q}{4}(1+2 \sqrt{2})\)
3 \(-\frac{Q}{2}(1+2 \sqrt{2})\)
4 \(\frac{Q}{2}(1+2 \sqrt{2})\)
Explanation:
b \[ \begin{aligned} & F_A=k \frac{Q^2}{a^2}, F_{\mathrm{C}:}=\frac{k Q^2}{a^2} \\ & F_D=\frac{k Q^2}{(a \sqrt{2})^2} \text { and } F_O=\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2} \\ & =\sqrt{F_A^2+F_{\mathrm{C}}^2}+F_D=\sqrt{2} \frac{k Q^2}{a^2}+\frac{k Q^2}{2 a^2}=\frac{k Q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right) \end{aligned} \] Force at \(B\) towards the centre \(=F_0=\frac{2 k Q q}{a^2}\) For equilibrium of charge at \(B, F_{p c}+F_D=F_o\) \[ \begin{aligned} & \Rightarrow \frac{K^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{2 K Q q}{a^2} \\ & \Rightarrow \mathrm{q}=\frac{Q}{4}(1+2 \sqrt{2}) \end{aligned} \]
266579
If graph between binding energy per nuclion with mass no. \(A\) is shown in figure there are 4 elements P, Q, R, S, shown on the graph then select the correct option :
1 P, Q may produce fussion reaction
2 P,Q may produce fission reaction
3 R,S, may produce fission reaction
4 Both \(A \& C\)
Explanation:
d \(\mathrm{P}, \mathrm{Q}\) nuclear fussion reaction shows R S nuclear fission reaction shows
**NCERT-450**
TEST SERIES (PHYSICS FST)
266580
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result or photon emission will be:
c According to law of conservation of momentum Momentum of photon \(=\) Momentum of atom or \[ \frac{h}{\lambda}=m v \]
**NCERT-427**
TEST SERIES (PHYSICS FST)
266581
In the figure (i) \& (ii) two vibrating air-colums of same length are shown. The ratio of frequencies in the two cases will be:
1 \(1: 2\)
2 \(3: 5\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
b \[ \begin{aligned} v=\frac{h}{\lambda m} & =\frac{h}{m} z^2 R\left[\frac{1}{t^2}-\frac{1}{5^2}\right] \\ & =\frac{h}{m} R\left[1-\frac{1}{25}\right]=\frac{24}{25} \frac{h R}{m} \end{aligned} \] In closed pipe. The frequencies are in odd ratio so \[ n_1: n_2: n_3: \ldots \ldots=1: 3: 5 \] Hence in case (i) \& (ii) the frequency ratio are 3 : 5
**NCERT -374**
TEST SERIES (PHYSICS FST)
266582
Four charges equal to \(-Q\) are placed at the four corners of a square and a charge q is placed at its centre. If the system is in equilibrium, the value of \(q\) is:
1 \(-\frac{Q}{4}(1+2 \sqrt{2})\)
2 \(\frac{Q}{4}(1+2 \sqrt{2})\)
3 \(-\frac{Q}{2}(1+2 \sqrt{2})\)
4 \(\frac{Q}{2}(1+2 \sqrt{2})\)
Explanation:
b \[ \begin{aligned} & F_A=k \frac{Q^2}{a^2}, F_{\mathrm{C}:}=\frac{k Q^2}{a^2} \\ & F_D=\frac{k Q^2}{(a \sqrt{2})^2} \text { and } F_O=\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2} \\ & =\sqrt{F_A^2+F_{\mathrm{C}}^2}+F_D=\sqrt{2} \frac{k Q^2}{a^2}+\frac{k Q^2}{2 a^2}=\frac{k Q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right) \end{aligned} \] Force at \(B\) towards the centre \(=F_0=\frac{2 k Q q}{a^2}\) For equilibrium of charge at \(B, F_{p c}+F_D=F_o\) \[ \begin{aligned} & \Rightarrow \frac{K^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{2 K Q q}{a^2} \\ & \Rightarrow \mathrm{q}=\frac{Q}{4}(1+2 \sqrt{2}) \end{aligned} \]
NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
266579
If graph between binding energy per nuclion with mass no. \(A\) is shown in figure there are 4 elements P, Q, R, S, shown on the graph then select the correct option :
1 P, Q may produce fussion reaction
2 P,Q may produce fission reaction
3 R,S, may produce fission reaction
4 Both \(A \& C\)
Explanation:
d \(\mathrm{P}, \mathrm{Q}\) nuclear fussion reaction shows R S nuclear fission reaction shows
**NCERT-450**
TEST SERIES (PHYSICS FST)
266580
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result or photon emission will be:
c According to law of conservation of momentum Momentum of photon \(=\) Momentum of atom or \[ \frac{h}{\lambda}=m v \]
**NCERT-427**
TEST SERIES (PHYSICS FST)
266581
In the figure (i) \& (ii) two vibrating air-colums of same length are shown. The ratio of frequencies in the two cases will be:
1 \(1: 2\)
2 \(3: 5\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
b \[ \begin{aligned} v=\frac{h}{\lambda m} & =\frac{h}{m} z^2 R\left[\frac{1}{t^2}-\frac{1}{5^2}\right] \\ & =\frac{h}{m} R\left[1-\frac{1}{25}\right]=\frac{24}{25} \frac{h R}{m} \end{aligned} \] In closed pipe. The frequencies are in odd ratio so \[ n_1: n_2: n_3: \ldots \ldots=1: 3: 5 \] Hence in case (i) \& (ii) the frequency ratio are 3 : 5
**NCERT -374**
TEST SERIES (PHYSICS FST)
266582
Four charges equal to \(-Q\) are placed at the four corners of a square and a charge q is placed at its centre. If the system is in equilibrium, the value of \(q\) is:
1 \(-\frac{Q}{4}(1+2 \sqrt{2})\)
2 \(\frac{Q}{4}(1+2 \sqrt{2})\)
3 \(-\frac{Q}{2}(1+2 \sqrt{2})\)
4 \(\frac{Q}{2}(1+2 \sqrt{2})\)
Explanation:
b \[ \begin{aligned} & F_A=k \frac{Q^2}{a^2}, F_{\mathrm{C}:}=\frac{k Q^2}{a^2} \\ & F_D=\frac{k Q^2}{(a \sqrt{2})^2} \text { and } F_O=\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2} \\ & =\sqrt{F_A^2+F_{\mathrm{C}}^2}+F_D=\sqrt{2} \frac{k Q^2}{a^2}+\frac{k Q^2}{2 a^2}=\frac{k Q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right) \end{aligned} \] Force at \(B\) towards the centre \(=F_0=\frac{2 k Q q}{a^2}\) For equilibrium of charge at \(B, F_{p c}+F_D=F_o\) \[ \begin{aligned} & \Rightarrow \frac{K^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{2 K Q q}{a^2} \\ & \Rightarrow \mathrm{q}=\frac{Q}{4}(1+2 \sqrt{2}) \end{aligned} \]
266579
If graph between binding energy per nuclion with mass no. \(A\) is shown in figure there are 4 elements P, Q, R, S, shown on the graph then select the correct option :
1 P, Q may produce fussion reaction
2 P,Q may produce fission reaction
3 R,S, may produce fission reaction
4 Both \(A \& C\)
Explanation:
d \(\mathrm{P}, \mathrm{Q}\) nuclear fussion reaction shows R S nuclear fission reaction shows
**NCERT-450**
TEST SERIES (PHYSICS FST)
266580
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result or photon emission will be:
c According to law of conservation of momentum Momentum of photon \(=\) Momentum of atom or \[ \frac{h}{\lambda}=m v \]
**NCERT-427**
TEST SERIES (PHYSICS FST)
266581
In the figure (i) \& (ii) two vibrating air-colums of same length are shown. The ratio of frequencies in the two cases will be:
1 \(1: 2\)
2 \(3: 5\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
b \[ \begin{aligned} v=\frac{h}{\lambda m} & =\frac{h}{m} z^2 R\left[\frac{1}{t^2}-\frac{1}{5^2}\right] \\ & =\frac{h}{m} R\left[1-\frac{1}{25}\right]=\frac{24}{25} \frac{h R}{m} \end{aligned} \] In closed pipe. The frequencies are in odd ratio so \[ n_1: n_2: n_3: \ldots \ldots=1: 3: 5 \] Hence in case (i) \& (ii) the frequency ratio are 3 : 5
**NCERT -374**
TEST SERIES (PHYSICS FST)
266582
Four charges equal to \(-Q\) are placed at the four corners of a square and a charge q is placed at its centre. If the system is in equilibrium, the value of \(q\) is:
1 \(-\frac{Q}{4}(1+2 \sqrt{2})\)
2 \(\frac{Q}{4}(1+2 \sqrt{2})\)
3 \(-\frac{Q}{2}(1+2 \sqrt{2})\)
4 \(\frac{Q}{2}(1+2 \sqrt{2})\)
Explanation:
b \[ \begin{aligned} & F_A=k \frac{Q^2}{a^2}, F_{\mathrm{C}:}=\frac{k Q^2}{a^2} \\ & F_D=\frac{k Q^2}{(a \sqrt{2})^2} \text { and } F_O=\frac{k Q q}{\left(\frac{a}{\sqrt{2}}\right)^2} \\ & =\sqrt{F_A^2+F_{\mathrm{C}}^2}+F_D=\sqrt{2} \frac{k Q^2}{a^2}+\frac{k Q^2}{2 a^2}=\frac{k Q^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right) \end{aligned} \] Force at \(B\) towards the centre \(=F_0=\frac{2 k Q q}{a^2}\) For equilibrium of charge at \(B, F_{p c}+F_D=F_o\) \[ \begin{aligned} & \Rightarrow \frac{K^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)=\frac{2 K Q q}{a^2} \\ & \Rightarrow \mathrm{q}=\frac{Q}{4}(1+2 \sqrt{2}) \end{aligned} \]
