151739
A cylindrical wire $P$ has resistance $10 \Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is
1 $10 \Omega$
2 $20 \Omega$
3 $5 \Omega$
4 $\frac{5}{2} \Omega$
Explanation:
B Given, Cylindrical wire $\mathrm{P}$ has resistance $=10 \Omega$ Length of wire $\mathrm{P}=l$ Diameter of wire $\mathrm{P}=\mathrm{d}$ Length of wire $\mathrm{Q}=\frac{l}{2}$ Diameter of wire $\mathrm{Q}=\frac{\mathrm{d}}{2}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\pi \mathrm{d}^{2} / 4\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ According to the questions, $\frac{\text { Resistance of } \mathrm{Q}}{\text { Resistance of } \mathrm{P}}=\frac{\frac{(l / 2)}{(\mathrm{d} / 2)^{2}}}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=\frac{2\left(\frac{l}{\mathrm{~d}^{2}}\right)}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=2$ Resistance of $\mathrm{Q}=$ Resistance of $\mathrm{P} \times 2$ $=10 \times 2$ $=20 \Omega$
TS- EAMCET-10.09.2020
Current Electricity
151740
A conducting wire of cross-sectional area $1 \mathrm{~cm}^{2}$ has $3 \times 10^{23}$ charge carriers per $\mathrm{m}^{3}$. If wire carries a current of $24 \mathrm{~mA}$, then drift velocity of carriers is
1 $5 \times 10^{-2} \mathrm{~m} / \mathrm{s}$
2 $0.5 \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
4 $5 \times 10^{-6} \mathrm{~m} / \mathrm{s}$
Explanation:
C Cross sectional area of conducting wire, $\mathrm{A}=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ Number of electron per $\mathrm{m}^{3}(\mathrm{n})=3 \times 10^{23} / \mathrm{m}^{3}$ Wire carries a current $(\mathrm{I})=24 \mathrm{~mA}=24 \times 10^{-3} \mathrm{~A}$ Electron charge $(e)=1.6 \times 10^{-19} \mathrm{C}$ Drift velocity $=\frac{\mathrm{I}}{\mathrm{ne} A}$ $=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 1 \times 10^{-4}}$ $=5 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
TS- EAMCET-11.09.2020
Current Electricity
151741
If resistivity of copper is $1.72 \times 10^{-8} \Omega-\mathrm{m}$ and number of free electrons in copper is $8.5 \times 10^{28} / \mathrm{m}^{3}$. Find the mobility.
151739
A cylindrical wire $P$ has resistance $10 \Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is
1 $10 \Omega$
2 $20 \Omega$
3 $5 \Omega$
4 $\frac{5}{2} \Omega$
Explanation:
B Given, Cylindrical wire $\mathrm{P}$ has resistance $=10 \Omega$ Length of wire $\mathrm{P}=l$ Diameter of wire $\mathrm{P}=\mathrm{d}$ Length of wire $\mathrm{Q}=\frac{l}{2}$ Diameter of wire $\mathrm{Q}=\frac{\mathrm{d}}{2}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\pi \mathrm{d}^{2} / 4\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ According to the questions, $\frac{\text { Resistance of } \mathrm{Q}}{\text { Resistance of } \mathrm{P}}=\frac{\frac{(l / 2)}{(\mathrm{d} / 2)^{2}}}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=\frac{2\left(\frac{l}{\mathrm{~d}^{2}}\right)}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=2$ Resistance of $\mathrm{Q}=$ Resistance of $\mathrm{P} \times 2$ $=10 \times 2$ $=20 \Omega$
TS- EAMCET-10.09.2020
Current Electricity
151740
A conducting wire of cross-sectional area $1 \mathrm{~cm}^{2}$ has $3 \times 10^{23}$ charge carriers per $\mathrm{m}^{3}$. If wire carries a current of $24 \mathrm{~mA}$, then drift velocity of carriers is
1 $5 \times 10^{-2} \mathrm{~m} / \mathrm{s}$
2 $0.5 \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
4 $5 \times 10^{-6} \mathrm{~m} / \mathrm{s}$
Explanation:
C Cross sectional area of conducting wire, $\mathrm{A}=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ Number of electron per $\mathrm{m}^{3}(\mathrm{n})=3 \times 10^{23} / \mathrm{m}^{3}$ Wire carries a current $(\mathrm{I})=24 \mathrm{~mA}=24 \times 10^{-3} \mathrm{~A}$ Electron charge $(e)=1.6 \times 10^{-19} \mathrm{C}$ Drift velocity $=\frac{\mathrm{I}}{\mathrm{ne} A}$ $=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 1 \times 10^{-4}}$ $=5 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
TS- EAMCET-11.09.2020
Current Electricity
151741
If resistivity of copper is $1.72 \times 10^{-8} \Omega-\mathrm{m}$ and number of free electrons in copper is $8.5 \times 10^{28} / \mathrm{m}^{3}$. Find the mobility.
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Current Electricity
151739
A cylindrical wire $P$ has resistance $10 \Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is
1 $10 \Omega$
2 $20 \Omega$
3 $5 \Omega$
4 $\frac{5}{2} \Omega$
Explanation:
B Given, Cylindrical wire $\mathrm{P}$ has resistance $=10 \Omega$ Length of wire $\mathrm{P}=l$ Diameter of wire $\mathrm{P}=\mathrm{d}$ Length of wire $\mathrm{Q}=\frac{l}{2}$ Diameter of wire $\mathrm{Q}=\frac{\mathrm{d}}{2}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\pi \mathrm{d}^{2} / 4\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ According to the questions, $\frac{\text { Resistance of } \mathrm{Q}}{\text { Resistance of } \mathrm{P}}=\frac{\frac{(l / 2)}{(\mathrm{d} / 2)^{2}}}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=\frac{2\left(\frac{l}{\mathrm{~d}^{2}}\right)}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=2$ Resistance of $\mathrm{Q}=$ Resistance of $\mathrm{P} \times 2$ $=10 \times 2$ $=20 \Omega$
TS- EAMCET-10.09.2020
Current Electricity
151740
A conducting wire of cross-sectional area $1 \mathrm{~cm}^{2}$ has $3 \times 10^{23}$ charge carriers per $\mathrm{m}^{3}$. If wire carries a current of $24 \mathrm{~mA}$, then drift velocity of carriers is
1 $5 \times 10^{-2} \mathrm{~m} / \mathrm{s}$
2 $0.5 \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
4 $5 \times 10^{-6} \mathrm{~m} / \mathrm{s}$
Explanation:
C Cross sectional area of conducting wire, $\mathrm{A}=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ Number of electron per $\mathrm{m}^{3}(\mathrm{n})=3 \times 10^{23} / \mathrm{m}^{3}$ Wire carries a current $(\mathrm{I})=24 \mathrm{~mA}=24 \times 10^{-3} \mathrm{~A}$ Electron charge $(e)=1.6 \times 10^{-19} \mathrm{C}$ Drift velocity $=\frac{\mathrm{I}}{\mathrm{ne} A}$ $=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 1 \times 10^{-4}}$ $=5 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
TS- EAMCET-11.09.2020
Current Electricity
151741
If resistivity of copper is $1.72 \times 10^{-8} \Omega-\mathrm{m}$ and number of free electrons in copper is $8.5 \times 10^{28} / \mathrm{m}^{3}$. Find the mobility.
151739
A cylindrical wire $P$ has resistance $10 \Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is
1 $10 \Omega$
2 $20 \Omega$
3 $5 \Omega$
4 $\frac{5}{2} \Omega$
Explanation:
B Given, Cylindrical wire $\mathrm{P}$ has resistance $=10 \Omega$ Length of wire $\mathrm{P}=l$ Diameter of wire $\mathrm{P}=\mathrm{d}$ Length of wire $\mathrm{Q}=\frac{l}{2}$ Diameter of wire $\mathrm{Q}=\frac{\mathrm{d}}{2}$ Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\pi \mathrm{d}^{2} / 4\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ According to the questions, $\frac{\text { Resistance of } \mathrm{Q}}{\text { Resistance of } \mathrm{P}}=\frac{\frac{(l / 2)}{(\mathrm{d} / 2)^{2}}}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=\frac{2\left(\frac{l}{\mathrm{~d}^{2}}\right)}{\left(\frac{l}{\mathrm{~d}^{2}}\right)}=2$ Resistance of $\mathrm{Q}=$ Resistance of $\mathrm{P} \times 2$ $=10 \times 2$ $=20 \Omega$
TS- EAMCET-10.09.2020
Current Electricity
151740
A conducting wire of cross-sectional area $1 \mathrm{~cm}^{2}$ has $3 \times 10^{23}$ charge carriers per $\mathrm{m}^{3}$. If wire carries a current of $24 \mathrm{~mA}$, then drift velocity of carriers is
1 $5 \times 10^{-2} \mathrm{~m} / \mathrm{s}$
2 $0.5 \mathrm{~m} / \mathrm{s}$
3 $5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$
4 $5 \times 10^{-6} \mathrm{~m} / \mathrm{s}$
Explanation:
C Cross sectional area of conducting wire, $\mathrm{A}=1 \mathrm{~cm}^{2}=1 \times 10^{-4} \mathrm{~m}^{2}$ Number of electron per $\mathrm{m}^{3}(\mathrm{n})=3 \times 10^{23} / \mathrm{m}^{3}$ Wire carries a current $(\mathrm{I})=24 \mathrm{~mA}=24 \times 10^{-3} \mathrm{~A}$ Electron charge $(e)=1.6 \times 10^{-19} \mathrm{C}$ Drift velocity $=\frac{\mathrm{I}}{\mathrm{ne} A}$ $=\frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 1 \times 10^{-4}}$ $=5 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
TS- EAMCET-11.09.2020
Current Electricity
151741
If resistivity of copper is $1.72 \times 10^{-8} \Omega-\mathrm{m}$ and number of free electrons in copper is $8.5 \times 10^{28} / \mathrm{m}^{3}$. Find the mobility.
