151952
Two copper wires have their masses in the ratio $2: 3$ and the lengths in the ratio $3: 4$, The ratio of their resistances is
1 $4: 9$
2 $27: 32$
3 $16: 9$
4 $27: 128$
5 $1: 2$
Explanation:
B Given, Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)=\frac{2}{3}$ length ratio $\left(\frac{l_{1}}{l_{2}}\right)=\frac{3}{4}$ mass $=$ density $\times$ volume $=$ density $\times$ area $\times$ length Area $=\frac{\operatorname{mass}(\mathrm{m})}{\operatorname{density}(\mathrm{d}) \times \text { length }(l)}$ $\mathrm{A}=\frac{\mathrm{m}}{\mathrm{d} \times l}$ $\mathrm{A} \propto \frac{\mathrm{m}}{l}$ $\text { Resistance } \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right) \times\left(\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}\right)$ $=\left(\frac{l_{1}}{l_{2}}\right)^{2} \times\left(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right)=\left(\frac{3}{4}\right)^{2} \times\left(\frac{3}{2}\right)=\frac{27}{32}$
Kerala CEE - 2009
Current Electricity
151954
In the electric circuit shown each cell has an emf of $2 \mathrm{~V}$ and internal resistance of $1 \Omega$. The external resistance is $2 \Omega$. The value of the current $I$ is : (in amperes):
1 2
2 1.25
3 0.4
4 1.2
5 0.8
Explanation:
D Given, Emf of each cell $(\mathrm{E})=2 \mathrm{~V}$ Internal resistance $(\mathrm{r})=1 \Omega$ External resistance $(\mathrm{R})=2 \Omega$ Apply KVL in the circuit $2-\mathrm{i}+2-\mathrm{i}+2-\mathrm{i}-2 \mathrm{i}=0$ $6-5 i=0$ $\mathrm{i}=\frac{6}{5}=1.2 \Omega$
151952
Two copper wires have their masses in the ratio $2: 3$ and the lengths in the ratio $3: 4$, The ratio of their resistances is
1 $4: 9$
2 $27: 32$
3 $16: 9$
4 $27: 128$
5 $1: 2$
Explanation:
B Given, Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)=\frac{2}{3}$ length ratio $\left(\frac{l_{1}}{l_{2}}\right)=\frac{3}{4}$ mass $=$ density $\times$ volume $=$ density $\times$ area $\times$ length Area $=\frac{\operatorname{mass}(\mathrm{m})}{\operatorname{density}(\mathrm{d}) \times \text { length }(l)}$ $\mathrm{A}=\frac{\mathrm{m}}{\mathrm{d} \times l}$ $\mathrm{A} \propto \frac{\mathrm{m}}{l}$ $\text { Resistance } \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right) \times\left(\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}\right)$ $=\left(\frac{l_{1}}{l_{2}}\right)^{2} \times\left(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right)=\left(\frac{3}{4}\right)^{2} \times\left(\frac{3}{2}\right)=\frac{27}{32}$
Kerala CEE - 2009
Current Electricity
151954
In the electric circuit shown each cell has an emf of $2 \mathrm{~V}$ and internal resistance of $1 \Omega$. The external resistance is $2 \Omega$. The value of the current $I$ is : (in amperes):
1 2
2 1.25
3 0.4
4 1.2
5 0.8
Explanation:
D Given, Emf of each cell $(\mathrm{E})=2 \mathrm{~V}$ Internal resistance $(\mathrm{r})=1 \Omega$ External resistance $(\mathrm{R})=2 \Omega$ Apply KVL in the circuit $2-\mathrm{i}+2-\mathrm{i}+2-\mathrm{i}-2 \mathrm{i}=0$ $6-5 i=0$ $\mathrm{i}=\frac{6}{5}=1.2 \Omega$
151952
Two copper wires have their masses in the ratio $2: 3$ and the lengths in the ratio $3: 4$, The ratio of their resistances is
1 $4: 9$
2 $27: 32$
3 $16: 9$
4 $27: 128$
5 $1: 2$
Explanation:
B Given, Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)=\frac{2}{3}$ length ratio $\left(\frac{l_{1}}{l_{2}}\right)=\frac{3}{4}$ mass $=$ density $\times$ volume $=$ density $\times$ area $\times$ length Area $=\frac{\operatorname{mass}(\mathrm{m})}{\operatorname{density}(\mathrm{d}) \times \text { length }(l)}$ $\mathrm{A}=\frac{\mathrm{m}}{\mathrm{d} \times l}$ $\mathrm{A} \propto \frac{\mathrm{m}}{l}$ $\text { Resistance } \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right) \times\left(\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}\right)$ $=\left(\frac{l_{1}}{l_{2}}\right)^{2} \times\left(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right)=\left(\frac{3}{4}\right)^{2} \times\left(\frac{3}{2}\right)=\frac{27}{32}$
Kerala CEE - 2009
Current Electricity
151954
In the electric circuit shown each cell has an emf of $2 \mathrm{~V}$ and internal resistance of $1 \Omega$. The external resistance is $2 \Omega$. The value of the current $I$ is : (in amperes):
1 2
2 1.25
3 0.4
4 1.2
5 0.8
Explanation:
D Given, Emf of each cell $(\mathrm{E})=2 \mathrm{~V}$ Internal resistance $(\mathrm{r})=1 \Omega$ External resistance $(\mathrm{R})=2 \Omega$ Apply KVL in the circuit $2-\mathrm{i}+2-\mathrm{i}+2-\mathrm{i}-2 \mathrm{i}=0$ $6-5 i=0$ $\mathrm{i}=\frac{6}{5}=1.2 \Omega$
151952
Two copper wires have their masses in the ratio $2: 3$ and the lengths in the ratio $3: 4$, The ratio of their resistances is
1 $4: 9$
2 $27: 32$
3 $16: 9$
4 $27: 128$
5 $1: 2$
Explanation:
B Given, Mass ratio $\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)=\frac{2}{3}$ length ratio $\left(\frac{l_{1}}{l_{2}}\right)=\frac{3}{4}$ mass $=$ density $\times$ volume $=$ density $\times$ area $\times$ length Area $=\frac{\operatorname{mass}(\mathrm{m})}{\operatorname{density}(\mathrm{d}) \times \text { length }(l)}$ $\mathrm{A}=\frac{\mathrm{m}}{\mathrm{d} \times l}$ $\mathrm{A} \propto \frac{\mathrm{m}}{l}$ $\text { Resistance } \quad \mathrm{R}=\rho \frac{l}{\mathrm{~A}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{l_{1}}{l_{2}}\right) \times\left(\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}\right)$ $=\left(\frac{l_{1}}{l_{2}}\right)^{2} \times\left(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right)=\left(\frac{3}{4}\right)^{2} \times\left(\frac{3}{2}\right)=\frac{27}{32}$
Kerala CEE - 2009
Current Electricity
151954
In the electric circuit shown each cell has an emf of $2 \mathrm{~V}$ and internal resistance of $1 \Omega$. The external resistance is $2 \Omega$. The value of the current $I$ is : (in amperes):
1 2
2 1.25
3 0.4
4 1.2
5 0.8
Explanation:
D Given, Emf of each cell $(\mathrm{E})=2 \mathrm{~V}$ Internal resistance $(\mathrm{r})=1 \Omega$ External resistance $(\mathrm{R})=2 \Omega$ Apply KVL in the circuit $2-\mathrm{i}+2-\mathrm{i}+2-\mathrm{i}-2 \mathrm{i}=0$ $6-5 i=0$ $\mathrm{i}=\frac{6}{5}=1.2 \Omega$
