NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155149
L-C-R series circuit contains a resistance of 10 $\Omega$ and self-inductance $0.4 \mathrm{H}$ connected in series with variable capacitor across $60 \mathrm{~V}$ and $50 \mathrm{~Hz}$ supply. The value of capacity at resonance will be $\pi^{2}=10$
155150
The correct graph of inductive reactance or capacitive reactance and frequency of the source of alternating signal is shown in
1 a
2 b
3 c
4 d
Explanation:
B $\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{fL}$ If $f$ increases then $X_{L}$ also increases linearly. $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ If $\mathrm{f}$ increases then $\mathrm{X}_{\mathrm{C}}$ decreases as in graph $\mathrm{b}$.
MHT-CET 2019
Alternating Current
155151
The inductance, capacitance and resistance are represented by ' $L$ ', ' $C$ ', ' $R$ ' respectively. Which one of the following term has the dimension of frequency?
1 $\frac{\mathrm{L}}{\mathrm{R}}$
2 $\frac{\mathrm{C}}{\mathrm{L}}$
3 $\mathrm{LC}$
4 $\frac{1}{\mathrm{RC}}$
Explanation:
D We know that, $\mathrm{V}=\mathrm{IR}, \quad \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ We know that, $\mathrm{V} =\frac{\mathrm{W}}{\mathrm{Q}}$ $\mathrm{V} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{AT}]}$ $\mathrm{V} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ $\therefore \quad \mathrm{R} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}{[\mathrm{A}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$ We know that, $\mathrm{C}=\mathrm{q} / \mathrm{V}$ $\mathrm{C}=\mathrm{It} / \mathrm{V}$ $\mathrm{C}=\frac{[\mathrm{AT}]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}$ Dimension of capacitance, $\mathrm{C}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$ Then dimension of $\frac{1}{R C}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ So, the dimension of $\frac{1}{\mathrm{RC}}$ is equal to dimension of frequency.
MHT-CET 2019
Alternating Current
155152
In a series $L C R$ circuit $R=300 \Omega, L=0.9 H, C$ $=2 \mu \mathrm{F}, \omega=1000 \mathrm{rad} / \mathrm{s}$. The impedance of the circuit is
1 $900 \Omega$
2 $500 \Omega$
3 $400 \Omega$
4 $1300 \Omega$
Explanation:
B Given that, $\mathrm{R}=300 \Omega, \mathrm{L}=0.9 \mathrm{H}, \mathrm{C}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$ $\omega=1000 \mathrm{rad} / \mathrm{sec}$ Impedance of the circuit, $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $Z=\sqrt{(300)^{2}+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^{2}}$ $Z=\sqrt{(300)^{2}+(400)^{2}}$ $Z=\sqrt{90000+160000}$ $Z=500 \Omega$
155149
L-C-R series circuit contains a resistance of 10 $\Omega$ and self-inductance $0.4 \mathrm{H}$ connected in series with variable capacitor across $60 \mathrm{~V}$ and $50 \mathrm{~Hz}$ supply. The value of capacity at resonance will be $\pi^{2}=10$
155150
The correct graph of inductive reactance or capacitive reactance and frequency of the source of alternating signal is shown in
1 a
2 b
3 c
4 d
Explanation:
B $\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{fL}$ If $f$ increases then $X_{L}$ also increases linearly. $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ If $\mathrm{f}$ increases then $\mathrm{X}_{\mathrm{C}}$ decreases as in graph $\mathrm{b}$.
MHT-CET 2019
Alternating Current
155151
The inductance, capacitance and resistance are represented by ' $L$ ', ' $C$ ', ' $R$ ' respectively. Which one of the following term has the dimension of frequency?
1 $\frac{\mathrm{L}}{\mathrm{R}}$
2 $\frac{\mathrm{C}}{\mathrm{L}}$
3 $\mathrm{LC}$
4 $\frac{1}{\mathrm{RC}}$
Explanation:
D We know that, $\mathrm{V}=\mathrm{IR}, \quad \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ We know that, $\mathrm{V} =\frac{\mathrm{W}}{\mathrm{Q}}$ $\mathrm{V} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{AT}]}$ $\mathrm{V} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ $\therefore \quad \mathrm{R} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}{[\mathrm{A}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$ We know that, $\mathrm{C}=\mathrm{q} / \mathrm{V}$ $\mathrm{C}=\mathrm{It} / \mathrm{V}$ $\mathrm{C}=\frac{[\mathrm{AT}]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}$ Dimension of capacitance, $\mathrm{C}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$ Then dimension of $\frac{1}{R C}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ So, the dimension of $\frac{1}{\mathrm{RC}}$ is equal to dimension of frequency.
MHT-CET 2019
Alternating Current
155152
In a series $L C R$ circuit $R=300 \Omega, L=0.9 H, C$ $=2 \mu \mathrm{F}, \omega=1000 \mathrm{rad} / \mathrm{s}$. The impedance of the circuit is
1 $900 \Omega$
2 $500 \Omega$
3 $400 \Omega$
4 $1300 \Omega$
Explanation:
B Given that, $\mathrm{R}=300 \Omega, \mathrm{L}=0.9 \mathrm{H}, \mathrm{C}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$ $\omega=1000 \mathrm{rad} / \mathrm{sec}$ Impedance of the circuit, $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $Z=\sqrt{(300)^{2}+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^{2}}$ $Z=\sqrt{(300)^{2}+(400)^{2}}$ $Z=\sqrt{90000+160000}$ $Z=500 \Omega$
155149
L-C-R series circuit contains a resistance of 10 $\Omega$ and self-inductance $0.4 \mathrm{H}$ connected in series with variable capacitor across $60 \mathrm{~V}$ and $50 \mathrm{~Hz}$ supply. The value of capacity at resonance will be $\pi^{2}=10$
155150
The correct graph of inductive reactance or capacitive reactance and frequency of the source of alternating signal is shown in
1 a
2 b
3 c
4 d
Explanation:
B $\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{fL}$ If $f$ increases then $X_{L}$ also increases linearly. $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ If $\mathrm{f}$ increases then $\mathrm{X}_{\mathrm{C}}$ decreases as in graph $\mathrm{b}$.
MHT-CET 2019
Alternating Current
155151
The inductance, capacitance and resistance are represented by ' $L$ ', ' $C$ ', ' $R$ ' respectively. Which one of the following term has the dimension of frequency?
1 $\frac{\mathrm{L}}{\mathrm{R}}$
2 $\frac{\mathrm{C}}{\mathrm{L}}$
3 $\mathrm{LC}$
4 $\frac{1}{\mathrm{RC}}$
Explanation:
D We know that, $\mathrm{V}=\mathrm{IR}, \quad \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ We know that, $\mathrm{V} =\frac{\mathrm{W}}{\mathrm{Q}}$ $\mathrm{V} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{AT}]}$ $\mathrm{V} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ $\therefore \quad \mathrm{R} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}{[\mathrm{A}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$ We know that, $\mathrm{C}=\mathrm{q} / \mathrm{V}$ $\mathrm{C}=\mathrm{It} / \mathrm{V}$ $\mathrm{C}=\frac{[\mathrm{AT}]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}$ Dimension of capacitance, $\mathrm{C}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$ Then dimension of $\frac{1}{R C}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ So, the dimension of $\frac{1}{\mathrm{RC}}$ is equal to dimension of frequency.
MHT-CET 2019
Alternating Current
155152
In a series $L C R$ circuit $R=300 \Omega, L=0.9 H, C$ $=2 \mu \mathrm{F}, \omega=1000 \mathrm{rad} / \mathrm{s}$. The impedance of the circuit is
1 $900 \Omega$
2 $500 \Omega$
3 $400 \Omega$
4 $1300 \Omega$
Explanation:
B Given that, $\mathrm{R}=300 \Omega, \mathrm{L}=0.9 \mathrm{H}, \mathrm{C}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$ $\omega=1000 \mathrm{rad} / \mathrm{sec}$ Impedance of the circuit, $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $Z=\sqrt{(300)^{2}+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^{2}}$ $Z=\sqrt{(300)^{2}+(400)^{2}}$ $Z=\sqrt{90000+160000}$ $Z=500 \Omega$
155149
L-C-R series circuit contains a resistance of 10 $\Omega$ and self-inductance $0.4 \mathrm{H}$ connected in series with variable capacitor across $60 \mathrm{~V}$ and $50 \mathrm{~Hz}$ supply. The value of capacity at resonance will be $\pi^{2}=10$
155150
The correct graph of inductive reactance or capacitive reactance and frequency of the source of alternating signal is shown in
1 a
2 b
3 c
4 d
Explanation:
B $\because \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \mathrm{fL}$ If $f$ increases then $X_{L}$ also increases linearly. $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{fC}}$ If $\mathrm{f}$ increases then $\mathrm{X}_{\mathrm{C}}$ decreases as in graph $\mathrm{b}$.
MHT-CET 2019
Alternating Current
155151
The inductance, capacitance and resistance are represented by ' $L$ ', ' $C$ ', ' $R$ ' respectively. Which one of the following term has the dimension of frequency?
1 $\frac{\mathrm{L}}{\mathrm{R}}$
2 $\frac{\mathrm{C}}{\mathrm{L}}$
3 $\mathrm{LC}$
4 $\frac{1}{\mathrm{RC}}$
Explanation:
D We know that, $\mathrm{V}=\mathrm{IR}, \quad \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$ We know that, $\mathrm{V} =\frac{\mathrm{W}}{\mathrm{Q}}$ $\mathrm{V} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{AT}]}$ $\mathrm{V} =\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ $\therefore \quad \mathrm{R} =\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}{[\mathrm{A}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$ We know that, $\mathrm{C}=\mathrm{q} / \mathrm{V}$ $\mathrm{C}=\mathrm{It} / \mathrm{V}$ $\mathrm{C}=\frac{[\mathrm{AT}]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]}$ Dimension of capacitance, $\mathrm{C}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$ Then dimension of $\frac{1}{R C}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ So, the dimension of $\frac{1}{\mathrm{RC}}$ is equal to dimension of frequency.
MHT-CET 2019
Alternating Current
155152
In a series $L C R$ circuit $R=300 \Omega, L=0.9 H, C$ $=2 \mu \mathrm{F}, \omega=1000 \mathrm{rad} / \mathrm{s}$. The impedance of the circuit is
1 $900 \Omega$
2 $500 \Omega$
3 $400 \Omega$
4 $1300 \Omega$
Explanation:
B Given that, $\mathrm{R}=300 \Omega, \mathrm{L}=0.9 \mathrm{H}, \mathrm{C}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$ $\omega=1000 \mathrm{rad} / \mathrm{sec}$ Impedance of the circuit, $Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $Z=\sqrt{(300)^{2}+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^{2}}$ $Z=\sqrt{(300)^{2}+(400)^{2}}$ $Z=\sqrt{90000+160000}$ $Z=500 \Omega$
