NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152900
The sensitivity of a galvanometer is $60 \mathrm{div} / \mathrm{A}$, When a shunt is used, its sensitivity becomes $10 \mathrm{div} / \mathrm{A}$. If the resistance of the galvanometer is $20 \Omega$, then the value of shunt used is
1 $4 \Omega$
2 $5 \Omega$
3 $20 \Omega$
4 $2 \Omega$
Explanation:
A Given, $I_{g}=10 \mathrm{div} / \mathrm{A}$ $I=60 \mathrm{div} / \mathrm{A}$ $\mathrm{G}=20 \Omega$ The Sensitivity of the galvanometer can be given as $\frac{I_{g}}{I}=\frac{S}{S+G}$ $\frac{10}{60}=\frac{S}{S+20}$ $\frac{1}{6}=\frac{S}{S+20}$ $S+20=6 S$ $5 S=20$ $S=4$
AP EAMCET-24.04.2018
Current Electricity
152901
A galvanometer has a coil of resistance $5 \Omega$ and requires $15 \mathrm{~mA}$ for full scale deflection. The shunt resistance needed to convert the galvanometer into an ammeter of range $0-1 \mathrm{~A}$ is
152902
A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement (s) is/are true?
1 The galvanometer resistance is infinite
2 The current through the galvanometer is 40 $\mathrm{mA}$.
3 After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor.
4 The galvanometer resistance is $150 \Omega$
Explanation:
B It is a case of balanced Wheatstone bridge $\frac{200}{300} =\frac{100}{\mathrm{G}}$ $\mathrm{G} =150 \Omega$ $\mathrm{R}_{\mathrm{eq}} =\frac{500 \times 250}{750}=\frac{500}{3} \Omega$ Now, the current through galvanometer $I =\frac{10}{250}=\frac{1}{25}$ $=0.04 \mathrm{~A}$ $\mathrm{I} =40 \mathrm{~mA}$
WB JEE 2018
Current Electricity
152903
In the circuit shown, the galvanometer of resistance $90 \Omega$ is shunted by a resistance of $r=$ $0.03 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ is nearly.
1 $9.97 \Omega$
2 $12 \Omega$
3 $5 \Omega$
4 $50 \Omega$
Explanation:
A Given, $\mathrm{R}_{\mathrm{G}}=90 \Omega, \mathrm{E}=10 \mathrm{~V}, \mathrm{I}=1 \mathrm{~A}$ Since, we know, when a galvanometer is shunted by a resistance $r$, its effective resistance, $\mathrm{R}_{\text {eff }}=\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}}=\frac{90 \times 0.03}{90+0.03} \approx 0.03 \Omega$ $\therefore \quad$ Net resistance of the circuit, $\mathrm{R}_{\mathrm{net}}=\mathrm{R}+\mathrm{R}_{\mathrm{eff}}$ $\mathrm{R}_{\mathrm{eff}}=\mathrm{R}+0.03$ $\because \quad \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{net}}} \Rightarrow \mathrm{R}_{\text {net }}=\frac{\mathrm{E}}{\mathrm{I}}$ $\mathrm{R}_{\text {net }}=\frac{10}{1}$ From equation (i), we get $\therefore \quad \mathrm{R}+0.03=10$ $\mathrm{R}=9.97 \Omega$
152900
The sensitivity of a galvanometer is $60 \mathrm{div} / \mathrm{A}$, When a shunt is used, its sensitivity becomes $10 \mathrm{div} / \mathrm{A}$. If the resistance of the galvanometer is $20 \Omega$, then the value of shunt used is
1 $4 \Omega$
2 $5 \Omega$
3 $20 \Omega$
4 $2 \Omega$
Explanation:
A Given, $I_{g}=10 \mathrm{div} / \mathrm{A}$ $I=60 \mathrm{div} / \mathrm{A}$ $\mathrm{G}=20 \Omega$ The Sensitivity of the galvanometer can be given as $\frac{I_{g}}{I}=\frac{S}{S+G}$ $\frac{10}{60}=\frac{S}{S+20}$ $\frac{1}{6}=\frac{S}{S+20}$ $S+20=6 S$ $5 S=20$ $S=4$
AP EAMCET-24.04.2018
Current Electricity
152901
A galvanometer has a coil of resistance $5 \Omega$ and requires $15 \mathrm{~mA}$ for full scale deflection. The shunt resistance needed to convert the galvanometer into an ammeter of range $0-1 \mathrm{~A}$ is
152902
A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement (s) is/are true?
1 The galvanometer resistance is infinite
2 The current through the galvanometer is 40 $\mathrm{mA}$.
3 After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor.
4 The galvanometer resistance is $150 \Omega$
Explanation:
B It is a case of balanced Wheatstone bridge $\frac{200}{300} =\frac{100}{\mathrm{G}}$ $\mathrm{G} =150 \Omega$ $\mathrm{R}_{\mathrm{eq}} =\frac{500 \times 250}{750}=\frac{500}{3} \Omega$ Now, the current through galvanometer $I =\frac{10}{250}=\frac{1}{25}$ $=0.04 \mathrm{~A}$ $\mathrm{I} =40 \mathrm{~mA}$
WB JEE 2018
Current Electricity
152903
In the circuit shown, the galvanometer of resistance $90 \Omega$ is shunted by a resistance of $r=$ $0.03 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ is nearly.
1 $9.97 \Omega$
2 $12 \Omega$
3 $5 \Omega$
4 $50 \Omega$
Explanation:
A Given, $\mathrm{R}_{\mathrm{G}}=90 \Omega, \mathrm{E}=10 \mathrm{~V}, \mathrm{I}=1 \mathrm{~A}$ Since, we know, when a galvanometer is shunted by a resistance $r$, its effective resistance, $\mathrm{R}_{\text {eff }}=\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}}=\frac{90 \times 0.03}{90+0.03} \approx 0.03 \Omega$ $\therefore \quad$ Net resistance of the circuit, $\mathrm{R}_{\mathrm{net}}=\mathrm{R}+\mathrm{R}_{\mathrm{eff}}$ $\mathrm{R}_{\mathrm{eff}}=\mathrm{R}+0.03$ $\because \quad \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{net}}} \Rightarrow \mathrm{R}_{\text {net }}=\frac{\mathrm{E}}{\mathrm{I}}$ $\mathrm{R}_{\text {net }}=\frac{10}{1}$ From equation (i), we get $\therefore \quad \mathrm{R}+0.03=10$ $\mathrm{R}=9.97 \Omega$
152900
The sensitivity of a galvanometer is $60 \mathrm{div} / \mathrm{A}$, When a shunt is used, its sensitivity becomes $10 \mathrm{div} / \mathrm{A}$. If the resistance of the galvanometer is $20 \Omega$, then the value of shunt used is
1 $4 \Omega$
2 $5 \Omega$
3 $20 \Omega$
4 $2 \Omega$
Explanation:
A Given, $I_{g}=10 \mathrm{div} / \mathrm{A}$ $I=60 \mathrm{div} / \mathrm{A}$ $\mathrm{G}=20 \Omega$ The Sensitivity of the galvanometer can be given as $\frac{I_{g}}{I}=\frac{S}{S+G}$ $\frac{10}{60}=\frac{S}{S+20}$ $\frac{1}{6}=\frac{S}{S+20}$ $S+20=6 S$ $5 S=20$ $S=4$
AP EAMCET-24.04.2018
Current Electricity
152901
A galvanometer has a coil of resistance $5 \Omega$ and requires $15 \mathrm{~mA}$ for full scale deflection. The shunt resistance needed to convert the galvanometer into an ammeter of range $0-1 \mathrm{~A}$ is
152902
A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement (s) is/are true?
1 The galvanometer resistance is infinite
2 The current through the galvanometer is 40 $\mathrm{mA}$.
3 After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor.
4 The galvanometer resistance is $150 \Omega$
Explanation:
B It is a case of balanced Wheatstone bridge $\frac{200}{300} =\frac{100}{\mathrm{G}}$ $\mathrm{G} =150 \Omega$ $\mathrm{R}_{\mathrm{eq}} =\frac{500 \times 250}{750}=\frac{500}{3} \Omega$ Now, the current through galvanometer $I =\frac{10}{250}=\frac{1}{25}$ $=0.04 \mathrm{~A}$ $\mathrm{I} =40 \mathrm{~mA}$
WB JEE 2018
Current Electricity
152903
In the circuit shown, the galvanometer of resistance $90 \Omega$ is shunted by a resistance of $r=$ $0.03 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ is nearly.
1 $9.97 \Omega$
2 $12 \Omega$
3 $5 \Omega$
4 $50 \Omega$
Explanation:
A Given, $\mathrm{R}_{\mathrm{G}}=90 \Omega, \mathrm{E}=10 \mathrm{~V}, \mathrm{I}=1 \mathrm{~A}$ Since, we know, when a galvanometer is shunted by a resistance $r$, its effective resistance, $\mathrm{R}_{\text {eff }}=\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}}=\frac{90 \times 0.03}{90+0.03} \approx 0.03 \Omega$ $\therefore \quad$ Net resistance of the circuit, $\mathrm{R}_{\mathrm{net}}=\mathrm{R}+\mathrm{R}_{\mathrm{eff}}$ $\mathrm{R}_{\mathrm{eff}}=\mathrm{R}+0.03$ $\because \quad \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{net}}} \Rightarrow \mathrm{R}_{\text {net }}=\frac{\mathrm{E}}{\mathrm{I}}$ $\mathrm{R}_{\text {net }}=\frac{10}{1}$ From equation (i), we get $\therefore \quad \mathrm{R}+0.03=10$ $\mathrm{R}=9.97 \Omega$
152900
The sensitivity of a galvanometer is $60 \mathrm{div} / \mathrm{A}$, When a shunt is used, its sensitivity becomes $10 \mathrm{div} / \mathrm{A}$. If the resistance of the galvanometer is $20 \Omega$, then the value of shunt used is
1 $4 \Omega$
2 $5 \Omega$
3 $20 \Omega$
4 $2 \Omega$
Explanation:
A Given, $I_{g}=10 \mathrm{div} / \mathrm{A}$ $I=60 \mathrm{div} / \mathrm{A}$ $\mathrm{G}=20 \Omega$ The Sensitivity of the galvanometer can be given as $\frac{I_{g}}{I}=\frac{S}{S+G}$ $\frac{10}{60}=\frac{S}{S+20}$ $\frac{1}{6}=\frac{S}{S+20}$ $S+20=6 S$ $5 S=20$ $S=4$
AP EAMCET-24.04.2018
Current Electricity
152901
A galvanometer has a coil of resistance $5 \Omega$ and requires $15 \mathrm{~mA}$ for full scale deflection. The shunt resistance needed to convert the galvanometer into an ammeter of range $0-1 \mathrm{~A}$ is
152902
A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement (s) is/are true?
1 The galvanometer resistance is infinite
2 The current through the galvanometer is 40 $\mathrm{mA}$.
3 After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor.
4 The galvanometer resistance is $150 \Omega$
Explanation:
B It is a case of balanced Wheatstone bridge $\frac{200}{300} =\frac{100}{\mathrm{G}}$ $\mathrm{G} =150 \Omega$ $\mathrm{R}_{\mathrm{eq}} =\frac{500 \times 250}{750}=\frac{500}{3} \Omega$ Now, the current through galvanometer $I =\frac{10}{250}=\frac{1}{25}$ $=0.04 \mathrm{~A}$ $\mathrm{I} =40 \mathrm{~mA}$
WB JEE 2018
Current Electricity
152903
In the circuit shown, the galvanometer of resistance $90 \Omega$ is shunted by a resistance of $r=$ $0.03 \Omega$. The current through $R$ is nearly $1 \mathrm{~A}$. The value of resistance $R$ is nearly.
1 $9.97 \Omega$
2 $12 \Omega$
3 $5 \Omega$
4 $50 \Omega$
Explanation:
A Given, $\mathrm{R}_{\mathrm{G}}=90 \Omega, \mathrm{E}=10 \mathrm{~V}, \mathrm{I}=1 \mathrm{~A}$ Since, we know, when a galvanometer is shunted by a resistance $r$, its effective resistance, $\mathrm{R}_{\text {eff }}=\frac{\mathrm{R}_{\mathrm{G}} \mathrm{r}}{\mathrm{R}_{\mathrm{G}}+\mathrm{r}}=\frac{90 \times 0.03}{90+0.03} \approx 0.03 \Omega$ $\therefore \quad$ Net resistance of the circuit, $\mathrm{R}_{\mathrm{net}}=\mathrm{R}+\mathrm{R}_{\mathrm{eff}}$ $\mathrm{R}_{\mathrm{eff}}=\mathrm{R}+0.03$ $\because \quad \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{net}}} \Rightarrow \mathrm{R}_{\text {net }}=\frac{\mathrm{E}}{\mathrm{I}}$ $\mathrm{R}_{\text {net }}=\frac{10}{1}$ From equation (i), we get $\therefore \quad \mathrm{R}+0.03=10$ $\mathrm{R}=9.97 \Omega$
