155172
The voltage time (V-t) graph for triangular wave having peak value $V_{0}$ is as shown in figure. The rms value of $\mathrm{V}$ in time interval from $t=0$ to $T / 4$ is $\frac{V_{0}}{\sqrt{x}}$ then find the value of $x$.
1 5
2 4
3 7
4 3
Explanation:
D Given, $\mathrm{V}_{\text {rms }}$ in $\mathrm{t}=0$ to $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ We know that, $\mathrm{V}=\frac{\mathrm{V}_{0}}{\mathrm{~T} / 4} \mathrm{t} \Rightarrow \mathrm{V}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \mathrm{t}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{ \lt \mathrm{V}^{2}>}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \sqrt{\left. \lt \mathrm{t}^{2}\right\rangle}$ $\mathrm{V}_{\mathrm{rms}}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}}\left\{\frac{\int_{0}^{\mathrm{T} / 4} \mathrm{t}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T} / 4} \mathrm{dt}}\right\}$ $\mathrm{V}_{\text {rms }}=\frac{\mathrm{V}_{0}}{\sqrt{3}}$ On comparing equation (i) \& (ii), we get - $\frac{\mathrm{V}_{0}}{\sqrt{3}}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ Hence, the value $\mathrm{x}=3$
AIIMS-27.05.2018(E)
Alternating Current
155173
In L-C-R circuit resonant frequency in $600 \mathrm{~Hz}$ and half power points are at 650 and $550 \mathrm{~Hz}$. The quality factor is
1 $\frac{1}{6}$
2 $\frac{1}{3}$
3 6
4 3
Explanation:
C Given, Resonance frequency $\left(\mathrm{f}_{0}\right)=600 \mathrm{~Hz}, \mathrm{f}_{2}=650 \mathrm{~Hz} \& \mathrm{f}_{1}=$ $650 \mathrm{~Hz}$ We know that, $\mathrm{B}=\frac{\mathrm{f}_{0}}{\mathrm{Q}}$ $\text { Where } \mathrm{B} =\text { Bandwidth }$ $\mathrm{Q} =\text { quality facror }$ $\text { Now, } \quad B=f_{2}-f_{1}=650-550=100 \mathrm{~Hz}$ $\mathrm{f}_{0}=600 \mathrm{~Hz}$ $\text { So, } \mathrm{Q}=\frac{600}{100}=6$
JCECE-2018
Alternating Current
155174
In the circuit shown, the symbols have their usual meanings. The cell has emf $E$. $X$ is initially joined to $Y$ for a long time. Then, $X$ is joined to $Z$. The maximum charge on $C$ at any later time will be
D Given circuit diagram - Case (I) - When switch is at ' $\mathrm{Y}$ ' for long times it means circuit goes in steady state condition and - $I=\frac{E}{R}$ Energy stored in inductor $\mathrm{U}=\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ Case (II) - When switch is connected to $\mathrm{Z}$ maximum energy in capacitor is - $\mathrm{U}=\frac{\mathrm{Q}_{\max }^{2}}{2 \mathrm{C}}$ $\mathrm{Q}_{\text {max }}^{2}=2 \mathrm{CE}$ $\mathrm{Q}_{\max }^{2}=2 \mathrm{C} \times \frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}^{2}=\mathrm{LC} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}=\frac{\mathrm{E}}{\mathrm{R}} \sqrt{\mathrm{LC}}$
Shift-1]
Alternating Current
155175
In the{AC}circuit shown, $E=E_{0} \sin (\omega t+\varphi) \text { and }$ Then, the box contains
1 Only C
2 $\mathrm{L}$ and $\mathrm{R}$ in series
3 C and R in series or L, C and R in series
4 Only R
Explanation:
C Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{i}=\mathrm{i}_{\mathrm{o}} \sin \left(\omega \mathrm{t}+\phi+\frac{\pi}{4}\right)$ As current leads voltage by $\pi / 4$, so circuit must be more capacitive than inductive. Hence it is either a $\mathrm{C}-\mathrm{R}$ combination or $\mathrm{L}-\mathrm{C}-\mathrm{R}$ combination.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155172
The voltage time (V-t) graph for triangular wave having peak value $V_{0}$ is as shown in figure. The rms value of $\mathrm{V}$ in time interval from $t=0$ to $T / 4$ is $\frac{V_{0}}{\sqrt{x}}$ then find the value of $x$.
1 5
2 4
3 7
4 3
Explanation:
D Given, $\mathrm{V}_{\text {rms }}$ in $\mathrm{t}=0$ to $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ We know that, $\mathrm{V}=\frac{\mathrm{V}_{0}}{\mathrm{~T} / 4} \mathrm{t} \Rightarrow \mathrm{V}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \mathrm{t}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{ \lt \mathrm{V}^{2}>}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \sqrt{\left. \lt \mathrm{t}^{2}\right\rangle}$ $\mathrm{V}_{\mathrm{rms}}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}}\left\{\frac{\int_{0}^{\mathrm{T} / 4} \mathrm{t}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T} / 4} \mathrm{dt}}\right\}$ $\mathrm{V}_{\text {rms }}=\frac{\mathrm{V}_{0}}{\sqrt{3}}$ On comparing equation (i) \& (ii), we get - $\frac{\mathrm{V}_{0}}{\sqrt{3}}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ Hence, the value $\mathrm{x}=3$
AIIMS-27.05.2018(E)
Alternating Current
155173
In L-C-R circuit resonant frequency in $600 \mathrm{~Hz}$ and half power points are at 650 and $550 \mathrm{~Hz}$. The quality factor is
1 $\frac{1}{6}$
2 $\frac{1}{3}$
3 6
4 3
Explanation:
C Given, Resonance frequency $\left(\mathrm{f}_{0}\right)=600 \mathrm{~Hz}, \mathrm{f}_{2}=650 \mathrm{~Hz} \& \mathrm{f}_{1}=$ $650 \mathrm{~Hz}$ We know that, $\mathrm{B}=\frac{\mathrm{f}_{0}}{\mathrm{Q}}$ $\text { Where } \mathrm{B} =\text { Bandwidth }$ $\mathrm{Q} =\text { quality facror }$ $\text { Now, } \quad B=f_{2}-f_{1}=650-550=100 \mathrm{~Hz}$ $\mathrm{f}_{0}=600 \mathrm{~Hz}$ $\text { So, } \mathrm{Q}=\frac{600}{100}=6$
JCECE-2018
Alternating Current
155174
In the circuit shown, the symbols have their usual meanings. The cell has emf $E$. $X$ is initially joined to $Y$ for a long time. Then, $X$ is joined to $Z$. The maximum charge on $C$ at any later time will be
D Given circuit diagram - Case (I) - When switch is at ' $\mathrm{Y}$ ' for long times it means circuit goes in steady state condition and - $I=\frac{E}{R}$ Energy stored in inductor $\mathrm{U}=\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ Case (II) - When switch is connected to $\mathrm{Z}$ maximum energy in capacitor is - $\mathrm{U}=\frac{\mathrm{Q}_{\max }^{2}}{2 \mathrm{C}}$ $\mathrm{Q}_{\text {max }}^{2}=2 \mathrm{CE}$ $\mathrm{Q}_{\max }^{2}=2 \mathrm{C} \times \frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}^{2}=\mathrm{LC} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}=\frac{\mathrm{E}}{\mathrm{R}} \sqrt{\mathrm{LC}}$
Shift-1]
Alternating Current
155175
In the{AC}circuit shown, $E=E_{0} \sin (\omega t+\varphi) \text { and }$ Then, the box contains
1 Only C
2 $\mathrm{L}$ and $\mathrm{R}$ in series
3 C and R in series or L, C and R in series
4 Only R
Explanation:
C Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{i}=\mathrm{i}_{\mathrm{o}} \sin \left(\omega \mathrm{t}+\phi+\frac{\pi}{4}\right)$ As current leads voltage by $\pi / 4$, so circuit must be more capacitive than inductive. Hence it is either a $\mathrm{C}-\mathrm{R}$ combination or $\mathrm{L}-\mathrm{C}-\mathrm{R}$ combination.
155172
The voltage time (V-t) graph for triangular wave having peak value $V_{0}$ is as shown in figure. The rms value of $\mathrm{V}$ in time interval from $t=0$ to $T / 4$ is $\frac{V_{0}}{\sqrt{x}}$ then find the value of $x$.
1 5
2 4
3 7
4 3
Explanation:
D Given, $\mathrm{V}_{\text {rms }}$ in $\mathrm{t}=0$ to $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ We know that, $\mathrm{V}=\frac{\mathrm{V}_{0}}{\mathrm{~T} / 4} \mathrm{t} \Rightarrow \mathrm{V}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \mathrm{t}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{ \lt \mathrm{V}^{2}>}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \sqrt{\left. \lt \mathrm{t}^{2}\right\rangle}$ $\mathrm{V}_{\mathrm{rms}}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}}\left\{\frac{\int_{0}^{\mathrm{T} / 4} \mathrm{t}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T} / 4} \mathrm{dt}}\right\}$ $\mathrm{V}_{\text {rms }}=\frac{\mathrm{V}_{0}}{\sqrt{3}}$ On comparing equation (i) \& (ii), we get - $\frac{\mathrm{V}_{0}}{\sqrt{3}}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ Hence, the value $\mathrm{x}=3$
AIIMS-27.05.2018(E)
Alternating Current
155173
In L-C-R circuit resonant frequency in $600 \mathrm{~Hz}$ and half power points are at 650 and $550 \mathrm{~Hz}$. The quality factor is
1 $\frac{1}{6}$
2 $\frac{1}{3}$
3 6
4 3
Explanation:
C Given, Resonance frequency $\left(\mathrm{f}_{0}\right)=600 \mathrm{~Hz}, \mathrm{f}_{2}=650 \mathrm{~Hz} \& \mathrm{f}_{1}=$ $650 \mathrm{~Hz}$ We know that, $\mathrm{B}=\frac{\mathrm{f}_{0}}{\mathrm{Q}}$ $\text { Where } \mathrm{B} =\text { Bandwidth }$ $\mathrm{Q} =\text { quality facror }$ $\text { Now, } \quad B=f_{2}-f_{1}=650-550=100 \mathrm{~Hz}$ $\mathrm{f}_{0}=600 \mathrm{~Hz}$ $\text { So, } \mathrm{Q}=\frac{600}{100}=6$
JCECE-2018
Alternating Current
155174
In the circuit shown, the symbols have their usual meanings. The cell has emf $E$. $X$ is initially joined to $Y$ for a long time. Then, $X$ is joined to $Z$. The maximum charge on $C$ at any later time will be
D Given circuit diagram - Case (I) - When switch is at ' $\mathrm{Y}$ ' for long times it means circuit goes in steady state condition and - $I=\frac{E}{R}$ Energy stored in inductor $\mathrm{U}=\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ Case (II) - When switch is connected to $\mathrm{Z}$ maximum energy in capacitor is - $\mathrm{U}=\frac{\mathrm{Q}_{\max }^{2}}{2 \mathrm{C}}$ $\mathrm{Q}_{\text {max }}^{2}=2 \mathrm{CE}$ $\mathrm{Q}_{\max }^{2}=2 \mathrm{C} \times \frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}^{2}=\mathrm{LC} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}=\frac{\mathrm{E}}{\mathrm{R}} \sqrt{\mathrm{LC}}$
Shift-1]
Alternating Current
155175
In the{AC}circuit shown, $E=E_{0} \sin (\omega t+\varphi) \text { and }$ Then, the box contains
1 Only C
2 $\mathrm{L}$ and $\mathrm{R}$ in series
3 C and R in series or L, C and R in series
4 Only R
Explanation:
C Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{i}=\mathrm{i}_{\mathrm{o}} \sin \left(\omega \mathrm{t}+\phi+\frac{\pi}{4}\right)$ As current leads voltage by $\pi / 4$, so circuit must be more capacitive than inductive. Hence it is either a $\mathrm{C}-\mathrm{R}$ combination or $\mathrm{L}-\mathrm{C}-\mathrm{R}$ combination.
155172
The voltage time (V-t) graph for triangular wave having peak value $V_{0}$ is as shown in figure. The rms value of $\mathrm{V}$ in time interval from $t=0$ to $T / 4$ is $\frac{V_{0}}{\sqrt{x}}$ then find the value of $x$.
1 5
2 4
3 7
4 3
Explanation:
D Given, $\mathrm{V}_{\text {rms }}$ in $\mathrm{t}=0$ to $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ We know that, $\mathrm{V}=\frac{\mathrm{V}_{0}}{\mathrm{~T} / 4} \mathrm{t} \Rightarrow \mathrm{V}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \mathrm{t}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{ \lt \mathrm{V}^{2}>}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}} \sqrt{\left. \lt \mathrm{t}^{2}\right\rangle}$ $\mathrm{V}_{\mathrm{rms}}=\frac{4 \mathrm{~V}_{0}}{\mathrm{~T}}\left\{\frac{\int_{0}^{\mathrm{T} / 4} \mathrm{t}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T} / 4} \mathrm{dt}}\right\}$ $\mathrm{V}_{\text {rms }}=\frac{\mathrm{V}_{0}}{\sqrt{3}}$ On comparing equation (i) \& (ii), we get - $\frac{\mathrm{V}_{0}}{\sqrt{3}}=\frac{\mathrm{V}_{0}}{\sqrt{\mathrm{x}}}$ Hence, the value $\mathrm{x}=3$
AIIMS-27.05.2018(E)
Alternating Current
155173
In L-C-R circuit resonant frequency in $600 \mathrm{~Hz}$ and half power points are at 650 and $550 \mathrm{~Hz}$. The quality factor is
1 $\frac{1}{6}$
2 $\frac{1}{3}$
3 6
4 3
Explanation:
C Given, Resonance frequency $\left(\mathrm{f}_{0}\right)=600 \mathrm{~Hz}, \mathrm{f}_{2}=650 \mathrm{~Hz} \& \mathrm{f}_{1}=$ $650 \mathrm{~Hz}$ We know that, $\mathrm{B}=\frac{\mathrm{f}_{0}}{\mathrm{Q}}$ $\text { Where } \mathrm{B} =\text { Bandwidth }$ $\mathrm{Q} =\text { quality facror }$ $\text { Now, } \quad B=f_{2}-f_{1}=650-550=100 \mathrm{~Hz}$ $\mathrm{f}_{0}=600 \mathrm{~Hz}$ $\text { So, } \mathrm{Q}=\frac{600}{100}=6$
JCECE-2018
Alternating Current
155174
In the circuit shown, the symbols have their usual meanings. The cell has emf $E$. $X$ is initially joined to $Y$ for a long time. Then, $X$ is joined to $Z$. The maximum charge on $C$ at any later time will be
D Given circuit diagram - Case (I) - When switch is at ' $\mathrm{Y}$ ' for long times it means circuit goes in steady state condition and - $I=\frac{E}{R}$ Energy stored in inductor $\mathrm{U}=\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ Case (II) - When switch is connected to $\mathrm{Z}$ maximum energy in capacitor is - $\mathrm{U}=\frac{\mathrm{Q}_{\max }^{2}}{2 \mathrm{C}}$ $\mathrm{Q}_{\text {max }}^{2}=2 \mathrm{CE}$ $\mathrm{Q}_{\max }^{2}=2 \mathrm{C} \times \frac{1}{2} \mathrm{~L} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}^{2}=\mathrm{LC} \frac{\mathrm{E}^{2}}{\mathrm{R}^{2}}$ $\mathrm{Q}=\frac{\mathrm{E}}{\mathrm{R}} \sqrt{\mathrm{LC}}$
Shift-1]
Alternating Current
155175
In the{AC}circuit shown, $E=E_{0} \sin (\omega t+\varphi) \text { and }$ Then, the box contains
1 Only C
2 $\mathrm{L}$ and $\mathrm{R}$ in series
3 C and R in series or L, C and R in series
4 Only R
Explanation:
C Given that, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{i}=\mathrm{i}_{\mathrm{o}} \sin \left(\omega \mathrm{t}+\phi+\frac{\pi}{4}\right)$ As current leads voltage by $\pi / 4$, so circuit must be more capacitive than inductive. Hence it is either a $\mathrm{C}-\mathrm{R}$ combination or $\mathrm{L}-\mathrm{C}-\mathrm{R}$ combination.
