151857
Resistance of a tungsten wire at $150^{\circ} \mathrm{C}$ is $133 \Omega$ its temperature coefficient of resistance is $0.0045^{\circ} \mathrm{C}^{-}$ ${ }^{1}$. The resistance of this wire at $500^{\circ} \mathrm{C}$ is
151858
There are four bulbs of power $100 \mathrm{~W}, 200 \mathrm{~W}$, $500 \mathrm{~W}$ and $1000 \mathrm{~W}$. Among these whose filament has high resistance? (Assuming, same voltage source)
1 $100 \mathrm{~W}$ bulb
2 $200 \mathrm{~W}$ bulb
3 $500 \mathrm{~W}$ bulb
4 $1000 \mathrm{~W}$ bulb
Explanation:
A Power rating of bulb $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Let, $\quad \mathrm{V}=200 \mathrm{~V}$ $\mathrm{R}_{100}=\frac{(200)^{2}}{100}=400 \mathrm{ohms}$ $\mathrm{R}_{200}=\frac{(200)^{2}}{200}=200 \mathrm{ohms}$ $\mathrm{R}_{500}=\frac{(200)^{2}}{500}=80 \mathrm{ohms}$ $\mathrm{R}_{1000}=\frac{(200)^{2}}{1000}=40 \text { ohms }$ $\mathrm{R}_{100}>\mathrm{R}_{200}>\mathrm{R}_{500}>\mathrm{R}_{1000}$ $\therefore 100 \mathrm{~W}$ bulb has high resistance,
TS- EAMCET-09.09.2020
Current Electricity
151859
A wire of resistance $5 \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is
1 $45 \Omega$
2 $54 \Omega$
3 $20 \Omega$
4 $5 \Omega$
Explanation:
A Given, $\mathrm{R}_{1}=5 \Omega$ $l_{1}=l$ $\mathrm{~A}_{1}=\mathrm{A}$ $l_{2}=2 l+l=3 l$ If $\mathrm{V}$ is the volume of wire of length and area of crosssection $\mathrm{A}$, then $\quad \mathrm{V}=\mathrm{A} l$ $\mathrm{A}=\frac{\mathrm{V}}{l}$ $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{(\mathrm{~V} / l)}=\frac{\rho l^{2}}{\mathrm{~V}}$ $\therefore \mathrm{R} \propto l^{2} \quad[\because \rho$ and $\mathrm{V}$ are constant $]$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{l_{2}^{2}}{l_{1}^{2}} \Rightarrow \frac{\mathrm{R}_{2}}{5}=\frac{(3 l)^{2}}{l^{2}}=9$ $\mathrm{R}_{2}=\mathrm{R}_{1} \times 9$ $\mathrm{R}_{2}=5 \times 9=45 \Omega$
151857
Resistance of a tungsten wire at $150^{\circ} \mathrm{C}$ is $133 \Omega$ its temperature coefficient of resistance is $0.0045^{\circ} \mathrm{C}^{-}$ ${ }^{1}$. The resistance of this wire at $500^{\circ} \mathrm{C}$ is
151858
There are four bulbs of power $100 \mathrm{~W}, 200 \mathrm{~W}$, $500 \mathrm{~W}$ and $1000 \mathrm{~W}$. Among these whose filament has high resistance? (Assuming, same voltage source)
1 $100 \mathrm{~W}$ bulb
2 $200 \mathrm{~W}$ bulb
3 $500 \mathrm{~W}$ bulb
4 $1000 \mathrm{~W}$ bulb
Explanation:
A Power rating of bulb $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Let, $\quad \mathrm{V}=200 \mathrm{~V}$ $\mathrm{R}_{100}=\frac{(200)^{2}}{100}=400 \mathrm{ohms}$ $\mathrm{R}_{200}=\frac{(200)^{2}}{200}=200 \mathrm{ohms}$ $\mathrm{R}_{500}=\frac{(200)^{2}}{500}=80 \mathrm{ohms}$ $\mathrm{R}_{1000}=\frac{(200)^{2}}{1000}=40 \text { ohms }$ $\mathrm{R}_{100}>\mathrm{R}_{200}>\mathrm{R}_{500}>\mathrm{R}_{1000}$ $\therefore 100 \mathrm{~W}$ bulb has high resistance,
TS- EAMCET-09.09.2020
Current Electricity
151859
A wire of resistance $5 \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is
1 $45 \Omega$
2 $54 \Omega$
3 $20 \Omega$
4 $5 \Omega$
Explanation:
A Given, $\mathrm{R}_{1}=5 \Omega$ $l_{1}=l$ $\mathrm{~A}_{1}=\mathrm{A}$ $l_{2}=2 l+l=3 l$ If $\mathrm{V}$ is the volume of wire of length and area of crosssection $\mathrm{A}$, then $\quad \mathrm{V}=\mathrm{A} l$ $\mathrm{A}=\frac{\mathrm{V}}{l}$ $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{(\mathrm{~V} / l)}=\frac{\rho l^{2}}{\mathrm{~V}}$ $\therefore \mathrm{R} \propto l^{2} \quad[\because \rho$ and $\mathrm{V}$ are constant $]$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{l_{2}^{2}}{l_{1}^{2}} \Rightarrow \frac{\mathrm{R}_{2}}{5}=\frac{(3 l)^{2}}{l^{2}}=9$ $\mathrm{R}_{2}=\mathrm{R}_{1} \times 9$ $\mathrm{R}_{2}=5 \times 9=45 \Omega$
151857
Resistance of a tungsten wire at $150^{\circ} \mathrm{C}$ is $133 \Omega$ its temperature coefficient of resistance is $0.0045^{\circ} \mathrm{C}^{-}$ ${ }^{1}$. The resistance of this wire at $500^{\circ} \mathrm{C}$ is
151858
There are four bulbs of power $100 \mathrm{~W}, 200 \mathrm{~W}$, $500 \mathrm{~W}$ and $1000 \mathrm{~W}$. Among these whose filament has high resistance? (Assuming, same voltage source)
1 $100 \mathrm{~W}$ bulb
2 $200 \mathrm{~W}$ bulb
3 $500 \mathrm{~W}$ bulb
4 $1000 \mathrm{~W}$ bulb
Explanation:
A Power rating of bulb $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Let, $\quad \mathrm{V}=200 \mathrm{~V}$ $\mathrm{R}_{100}=\frac{(200)^{2}}{100}=400 \mathrm{ohms}$ $\mathrm{R}_{200}=\frac{(200)^{2}}{200}=200 \mathrm{ohms}$ $\mathrm{R}_{500}=\frac{(200)^{2}}{500}=80 \mathrm{ohms}$ $\mathrm{R}_{1000}=\frac{(200)^{2}}{1000}=40 \text { ohms }$ $\mathrm{R}_{100}>\mathrm{R}_{200}>\mathrm{R}_{500}>\mathrm{R}_{1000}$ $\therefore 100 \mathrm{~W}$ bulb has high resistance,
TS- EAMCET-09.09.2020
Current Electricity
151859
A wire of resistance $5 \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is
1 $45 \Omega$
2 $54 \Omega$
3 $20 \Omega$
4 $5 \Omega$
Explanation:
A Given, $\mathrm{R}_{1}=5 \Omega$ $l_{1}=l$ $\mathrm{~A}_{1}=\mathrm{A}$ $l_{2}=2 l+l=3 l$ If $\mathrm{V}$ is the volume of wire of length and area of crosssection $\mathrm{A}$, then $\quad \mathrm{V}=\mathrm{A} l$ $\mathrm{A}=\frac{\mathrm{V}}{l}$ $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{(\mathrm{~V} / l)}=\frac{\rho l^{2}}{\mathrm{~V}}$ $\therefore \mathrm{R} \propto l^{2} \quad[\because \rho$ and $\mathrm{V}$ are constant $]$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{l_{2}^{2}}{l_{1}^{2}} \Rightarrow \frac{\mathrm{R}_{2}}{5}=\frac{(3 l)^{2}}{l^{2}}=9$ $\mathrm{R}_{2}=\mathrm{R}_{1} \times 9$ $\mathrm{R}_{2}=5 \times 9=45 \Omega$
151857
Resistance of a tungsten wire at $150^{\circ} \mathrm{C}$ is $133 \Omega$ its temperature coefficient of resistance is $0.0045^{\circ} \mathrm{C}^{-}$ ${ }^{1}$. The resistance of this wire at $500^{\circ} \mathrm{C}$ is
151858
There are four bulbs of power $100 \mathrm{~W}, 200 \mathrm{~W}$, $500 \mathrm{~W}$ and $1000 \mathrm{~W}$. Among these whose filament has high resistance? (Assuming, same voltage source)
1 $100 \mathrm{~W}$ bulb
2 $200 \mathrm{~W}$ bulb
3 $500 \mathrm{~W}$ bulb
4 $1000 \mathrm{~W}$ bulb
Explanation:
A Power rating of bulb $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Let, $\quad \mathrm{V}=200 \mathrm{~V}$ $\mathrm{R}_{100}=\frac{(200)^{2}}{100}=400 \mathrm{ohms}$ $\mathrm{R}_{200}=\frac{(200)^{2}}{200}=200 \mathrm{ohms}$ $\mathrm{R}_{500}=\frac{(200)^{2}}{500}=80 \mathrm{ohms}$ $\mathrm{R}_{1000}=\frac{(200)^{2}}{1000}=40 \text { ohms }$ $\mathrm{R}_{100}>\mathrm{R}_{200}>\mathrm{R}_{500}>\mathrm{R}_{1000}$ $\therefore 100 \mathrm{~W}$ bulb has high resistance,
TS- EAMCET-09.09.2020
Current Electricity
151859
A wire of resistance $5 \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is
1 $45 \Omega$
2 $54 \Omega$
3 $20 \Omega$
4 $5 \Omega$
Explanation:
A Given, $\mathrm{R}_{1}=5 \Omega$ $l_{1}=l$ $\mathrm{~A}_{1}=\mathrm{A}$ $l_{2}=2 l+l=3 l$ If $\mathrm{V}$ is the volume of wire of length and area of crosssection $\mathrm{A}$, then $\quad \mathrm{V}=\mathrm{A} l$ $\mathrm{A}=\frac{\mathrm{V}}{l}$ $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{(\mathrm{~V} / l)}=\frac{\rho l^{2}}{\mathrm{~V}}$ $\therefore \mathrm{R} \propto l^{2} \quad[\because \rho$ and $\mathrm{V}$ are constant $]$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{l_{2}^{2}}{l_{1}^{2}} \Rightarrow \frac{\mathrm{R}_{2}}{5}=\frac{(3 l)^{2}}{l^{2}}=9$ $\mathrm{R}_{2}=\mathrm{R}_{1} \times 9$ $\mathrm{R}_{2}=5 \times 9=45 \Omega$
