NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152801
In a meter bridge experiment, the balance point from left is $37.5 \mathrm{~cm}$. The ratio of the right gap resistance to the left gap resistance is:
1 $\frac{5}{3}$
2 $\frac{8}{5}$
3 $\frac{4}{5}$
4 $\frac{3}{2}$
Explanation:
A Given, Length of meter bridge, $l=37.5 \mathrm{~cm}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}$ $=\frac{37.5}{100-37.5}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{37.5}{62.5}=\frac{15}{25}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{3}{5}$ $\frac{\mathrm{S}}{\mathrm{R}}=\frac{5}{3}$
TS EAMCET 05.08.2021
Current Electricity
152803
For the which value of Resistance $R=$ when galvanometer shows zero deflection for following below electrical circuit.
1 $100 \Omega$
2 $300 \Omega$
3 $200 \Omega$
4 $400 \Omega$
Explanation:
A Current through $(\mathrm{R})=\frac{12}{500+\mathrm{R}}$ $\text { Voltage across }(R)=\frac{12 R}{500+R}$ Since, Galvanometer show zero deflection, $\frac{12 R}{500+R}=2$ $1000+2 R=12 R$ $R=100 \Omega$
GUJCET-PCE- 2021
Current Electricity
152804 How much resistance must be put in parallel to the resistance $S$ to balance the above bridge ?
152805
In a potentiometer circuit, a cell of emf $1.5 \mathrm{~V}$ gives balance point at $36 \mathrm{~cm}$ length of wire. If another cell of emf $2.5 \mathrm{~V}$ replaces the first cell, then at what length of the wire, the balance point occurs?
1 $60 \mathrm{~cm}$
2 $21.6 \mathrm{~cm}$
3 $64 \mathrm{~cm}$
4 $62 \mathrm{~cm}$
Explanation:
A Given that, Primary emf of cell $\left(\mathrm{E}_{1}\right)=1.5 \mathrm{~V}$ Secondary emf of cell $\left(\mathrm{E}_{2}\right)=2.5 \mathrm{~V}$ Balancing length $\left(l_{1}\right)=36 \mathrm{~cm}$ Equating emf of two cells- $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\frac{1.5}{2.5}=\frac{36}{l_{2}}$ $l_{2}=36 \times \frac{5}{3}=60 \mathrm{~cm} .$
152801
In a meter bridge experiment, the balance point from left is $37.5 \mathrm{~cm}$. The ratio of the right gap resistance to the left gap resistance is:
1 $\frac{5}{3}$
2 $\frac{8}{5}$
3 $\frac{4}{5}$
4 $\frac{3}{2}$
Explanation:
A Given, Length of meter bridge, $l=37.5 \mathrm{~cm}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}$ $=\frac{37.5}{100-37.5}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{37.5}{62.5}=\frac{15}{25}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{3}{5}$ $\frac{\mathrm{S}}{\mathrm{R}}=\frac{5}{3}$
TS EAMCET 05.08.2021
Current Electricity
152803
For the which value of Resistance $R=$ when galvanometer shows zero deflection for following below electrical circuit.
1 $100 \Omega$
2 $300 \Omega$
3 $200 \Omega$
4 $400 \Omega$
Explanation:
A Current through $(\mathrm{R})=\frac{12}{500+\mathrm{R}}$ $\text { Voltage across }(R)=\frac{12 R}{500+R}$ Since, Galvanometer show zero deflection, $\frac{12 R}{500+R}=2$ $1000+2 R=12 R$ $R=100 \Omega$
GUJCET-PCE- 2021
Current Electricity
152804 How much resistance must be put in parallel to the resistance $S$ to balance the above bridge ?
152805
In a potentiometer circuit, a cell of emf $1.5 \mathrm{~V}$ gives balance point at $36 \mathrm{~cm}$ length of wire. If another cell of emf $2.5 \mathrm{~V}$ replaces the first cell, then at what length of the wire, the balance point occurs?
1 $60 \mathrm{~cm}$
2 $21.6 \mathrm{~cm}$
3 $64 \mathrm{~cm}$
4 $62 \mathrm{~cm}$
Explanation:
A Given that, Primary emf of cell $\left(\mathrm{E}_{1}\right)=1.5 \mathrm{~V}$ Secondary emf of cell $\left(\mathrm{E}_{2}\right)=2.5 \mathrm{~V}$ Balancing length $\left(l_{1}\right)=36 \mathrm{~cm}$ Equating emf of two cells- $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\frac{1.5}{2.5}=\frac{36}{l_{2}}$ $l_{2}=36 \times \frac{5}{3}=60 \mathrm{~cm} .$
152801
In a meter bridge experiment, the balance point from left is $37.5 \mathrm{~cm}$. The ratio of the right gap resistance to the left gap resistance is:
1 $\frac{5}{3}$
2 $\frac{8}{5}$
3 $\frac{4}{5}$
4 $\frac{3}{2}$
Explanation:
A Given, Length of meter bridge, $l=37.5 \mathrm{~cm}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}$ $=\frac{37.5}{100-37.5}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{37.5}{62.5}=\frac{15}{25}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{3}{5}$ $\frac{\mathrm{S}}{\mathrm{R}}=\frac{5}{3}$
TS EAMCET 05.08.2021
Current Electricity
152803
For the which value of Resistance $R=$ when galvanometer shows zero deflection for following below electrical circuit.
1 $100 \Omega$
2 $300 \Omega$
3 $200 \Omega$
4 $400 \Omega$
Explanation:
A Current through $(\mathrm{R})=\frac{12}{500+\mathrm{R}}$ $\text { Voltage across }(R)=\frac{12 R}{500+R}$ Since, Galvanometer show zero deflection, $\frac{12 R}{500+R}=2$ $1000+2 R=12 R$ $R=100 \Omega$
GUJCET-PCE- 2021
Current Electricity
152804 How much resistance must be put in parallel to the resistance $S$ to balance the above bridge ?
152805
In a potentiometer circuit, a cell of emf $1.5 \mathrm{~V}$ gives balance point at $36 \mathrm{~cm}$ length of wire. If another cell of emf $2.5 \mathrm{~V}$ replaces the first cell, then at what length of the wire, the balance point occurs?
1 $60 \mathrm{~cm}$
2 $21.6 \mathrm{~cm}$
3 $64 \mathrm{~cm}$
4 $62 \mathrm{~cm}$
Explanation:
A Given that, Primary emf of cell $\left(\mathrm{E}_{1}\right)=1.5 \mathrm{~V}$ Secondary emf of cell $\left(\mathrm{E}_{2}\right)=2.5 \mathrm{~V}$ Balancing length $\left(l_{1}\right)=36 \mathrm{~cm}$ Equating emf of two cells- $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\frac{1.5}{2.5}=\frac{36}{l_{2}}$ $l_{2}=36 \times \frac{5}{3}=60 \mathrm{~cm} .$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152801
In a meter bridge experiment, the balance point from left is $37.5 \mathrm{~cm}$. The ratio of the right gap resistance to the left gap resistance is:
1 $\frac{5}{3}$
2 $\frac{8}{5}$
3 $\frac{4}{5}$
4 $\frac{3}{2}$
Explanation:
A Given, Length of meter bridge, $l=37.5 \mathrm{~cm}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{l}{100-l}$ $=\frac{37.5}{100-37.5}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{37.5}{62.5}=\frac{15}{25}$ $\frac{\mathrm{R}}{\mathrm{S}}=\frac{3}{5}$ $\frac{\mathrm{S}}{\mathrm{R}}=\frac{5}{3}$
TS EAMCET 05.08.2021
Current Electricity
152803
For the which value of Resistance $R=$ when galvanometer shows zero deflection for following below electrical circuit.
1 $100 \Omega$
2 $300 \Omega$
3 $200 \Omega$
4 $400 \Omega$
Explanation:
A Current through $(\mathrm{R})=\frac{12}{500+\mathrm{R}}$ $\text { Voltage across }(R)=\frac{12 R}{500+R}$ Since, Galvanometer show zero deflection, $\frac{12 R}{500+R}=2$ $1000+2 R=12 R$ $R=100 \Omega$
GUJCET-PCE- 2021
Current Electricity
152804 How much resistance must be put in parallel to the resistance $S$ to balance the above bridge ?
152805
In a potentiometer circuit, a cell of emf $1.5 \mathrm{~V}$ gives balance point at $36 \mathrm{~cm}$ length of wire. If another cell of emf $2.5 \mathrm{~V}$ replaces the first cell, then at what length of the wire, the balance point occurs?
1 $60 \mathrm{~cm}$
2 $21.6 \mathrm{~cm}$
3 $64 \mathrm{~cm}$
4 $62 \mathrm{~cm}$
Explanation:
A Given that, Primary emf of cell $\left(\mathrm{E}_{1}\right)=1.5 \mathrm{~V}$ Secondary emf of cell $\left(\mathrm{E}_{2}\right)=2.5 \mathrm{~V}$ Balancing length $\left(l_{1}\right)=36 \mathrm{~cm}$ Equating emf of two cells- $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}$ $\frac{1.5}{2.5}=\frac{36}{l_{2}}$ $l_{2}=36 \times \frac{5}{3}=60 \mathrm{~cm} .$
