151942
Two wires have lengths, diameters and specific resistance all in the ratio of $1: 2$. The resistance of the first wire is $10 \mathrm{ohm}$. Resistance of the second wire in ohm will be
151943
A light emitting diode (LED) has a voltage drop of $2 \mathrm{~V}$ across it and passes a current of $10 \mathrm{~mA}$. When it operates with a $6 \mathrm{~V}$ battery through a limiting resistor $R$, the value of $R$ is:
1 $40 \mathrm{k} \Omega$
2 $4 \mathrm{k} \Omega$
3 $200 \mathrm{k} \Omega$
4 $400 \Omega$
Explanation:
D Given, Current $(\mathrm{I})=10 \mathrm{~mA}$ $=0.01 \mathrm{~A}$ Voltage $\left(\mathrm{V}_{1}\right)=2 \mathrm{~V}$ Battery Voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Total Voltage $\left(\mathrm{V}_{\mathrm{T}}\right)=\mathrm{V}_{2}-\mathrm{V}_{1}=6-2$ By ohm's law, $=4 \mathrm{~V}$ $\mathrm{V}=\mathrm{IR}$ $4=0.01 \times R$ $\Rightarrow \quad \mathrm{R}=\frac{4}{0.01}=400 \Omega$
UPSEE - 2006
Current Electricity
151945
The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance?
A We know that, Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\frac{\pi d^{2}}{4}\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ For wire $\mathrm{A}=\frac{l}{\mathrm{~d}^{2}}=\frac{50}{(0.05)^{2}}=20000$ For wire $\mathrm{B}=\frac{l}{\mathrm{~d}^{2}}=\frac{100}{(0.1)^{2}}=10000$ For wire $\mathrm{C}=\frac{l}{\mathrm{~d}^{2}}=\frac{200}{(0.2)^{2}}=500$ For wire $\mathrm{D}=\frac{l}{\mathrm{~d}^{2}}=\frac{300}{(0.3)^{2}}=3333.33$ For highest resistance, the value of $\frac{l}{\mathrm{~d}^{2}}$ should be maximum.
UPSEE - 2004
Current Electricity
151946
Pick out the wrong statement from the following
1 The SI unit of conductance is mho
2 Conductance of a conductor decreases with increase in temperature
3 If the radius of a metallic wire is doubled, its resistance becomes $\left(\frac{1}{4}\right)^{\text {th }}$ of original resistance
4 If the length of the metallic wire is doubled, its resistivity remains uncharged
5 The relation between voltage and current for a non-ohmic conductor is linear
Explanation:
E (i) SI unit of the conductance is mho or siemens. (ii) The relation between resistance $\mathrm{R}$ and temperature $\mathrm{T}$ is as follows. $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)\right]$ Thus, as the resistance is increasing with increase in temperature, so the conductance will decrease with increase in temperature. (iii) The resistance is inversely proportional to square of radius of the wire. So, when the radius is doubled, the resistance will be $1 / 4^{\text {th }}$ of the original resistance. (iv) Resistivity is a constant term for a given material. It doesn't change by changing the length and crosssectional area of the material. (v) The voltage-current relation for ohmic conductor is linear but for a non-ohmic conductor it is non-linear.
Kerala CEE- 2013
Current Electricity
151949
The resistance of a $10 \mathrm{~m}$ long wire is $10 \Omega$. Its length is increased by $25 \%$ by stretching the wire uniformly. Then the resistance of the wire will be
151942
Two wires have lengths, diameters and specific resistance all in the ratio of $1: 2$. The resistance of the first wire is $10 \mathrm{ohm}$. Resistance of the second wire in ohm will be
151943
A light emitting diode (LED) has a voltage drop of $2 \mathrm{~V}$ across it and passes a current of $10 \mathrm{~mA}$. When it operates with a $6 \mathrm{~V}$ battery through a limiting resistor $R$, the value of $R$ is:
1 $40 \mathrm{k} \Omega$
2 $4 \mathrm{k} \Omega$
3 $200 \mathrm{k} \Omega$
4 $400 \Omega$
Explanation:
D Given, Current $(\mathrm{I})=10 \mathrm{~mA}$ $=0.01 \mathrm{~A}$ Voltage $\left(\mathrm{V}_{1}\right)=2 \mathrm{~V}$ Battery Voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Total Voltage $\left(\mathrm{V}_{\mathrm{T}}\right)=\mathrm{V}_{2}-\mathrm{V}_{1}=6-2$ By ohm's law, $=4 \mathrm{~V}$ $\mathrm{V}=\mathrm{IR}$ $4=0.01 \times R$ $\Rightarrow \quad \mathrm{R}=\frac{4}{0.01}=400 \Omega$
UPSEE - 2006
Current Electricity
151945
The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance?
A We know that, Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\frac{\pi d^{2}}{4}\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ For wire $\mathrm{A}=\frac{l}{\mathrm{~d}^{2}}=\frac{50}{(0.05)^{2}}=20000$ For wire $\mathrm{B}=\frac{l}{\mathrm{~d}^{2}}=\frac{100}{(0.1)^{2}}=10000$ For wire $\mathrm{C}=\frac{l}{\mathrm{~d}^{2}}=\frac{200}{(0.2)^{2}}=500$ For wire $\mathrm{D}=\frac{l}{\mathrm{~d}^{2}}=\frac{300}{(0.3)^{2}}=3333.33$ For highest resistance, the value of $\frac{l}{\mathrm{~d}^{2}}$ should be maximum.
UPSEE - 2004
Current Electricity
151946
Pick out the wrong statement from the following
1 The SI unit of conductance is mho
2 Conductance of a conductor decreases with increase in temperature
3 If the radius of a metallic wire is doubled, its resistance becomes $\left(\frac{1}{4}\right)^{\text {th }}$ of original resistance
4 If the length of the metallic wire is doubled, its resistivity remains uncharged
5 The relation between voltage and current for a non-ohmic conductor is linear
Explanation:
E (i) SI unit of the conductance is mho or siemens. (ii) The relation between resistance $\mathrm{R}$ and temperature $\mathrm{T}$ is as follows. $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)\right]$ Thus, as the resistance is increasing with increase in temperature, so the conductance will decrease with increase in temperature. (iii) The resistance is inversely proportional to square of radius of the wire. So, when the radius is doubled, the resistance will be $1 / 4^{\text {th }}$ of the original resistance. (iv) Resistivity is a constant term for a given material. It doesn't change by changing the length and crosssectional area of the material. (v) The voltage-current relation for ohmic conductor is linear but for a non-ohmic conductor it is non-linear.
Kerala CEE- 2013
Current Electricity
151949
The resistance of a $10 \mathrm{~m}$ long wire is $10 \Omega$. Its length is increased by $25 \%$ by stretching the wire uniformly. Then the resistance of the wire will be
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
151942
Two wires have lengths, diameters and specific resistance all in the ratio of $1: 2$. The resistance of the first wire is $10 \mathrm{ohm}$. Resistance of the second wire in ohm will be
151943
A light emitting diode (LED) has a voltage drop of $2 \mathrm{~V}$ across it and passes a current of $10 \mathrm{~mA}$. When it operates with a $6 \mathrm{~V}$ battery through a limiting resistor $R$, the value of $R$ is:
1 $40 \mathrm{k} \Omega$
2 $4 \mathrm{k} \Omega$
3 $200 \mathrm{k} \Omega$
4 $400 \Omega$
Explanation:
D Given, Current $(\mathrm{I})=10 \mathrm{~mA}$ $=0.01 \mathrm{~A}$ Voltage $\left(\mathrm{V}_{1}\right)=2 \mathrm{~V}$ Battery Voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Total Voltage $\left(\mathrm{V}_{\mathrm{T}}\right)=\mathrm{V}_{2}-\mathrm{V}_{1}=6-2$ By ohm's law, $=4 \mathrm{~V}$ $\mathrm{V}=\mathrm{IR}$ $4=0.01 \times R$ $\Rightarrow \quad \mathrm{R}=\frac{4}{0.01}=400 \Omega$
UPSEE - 2006
Current Electricity
151945
The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance?
A We know that, Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\frac{\pi d^{2}}{4}\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ For wire $\mathrm{A}=\frac{l}{\mathrm{~d}^{2}}=\frac{50}{(0.05)^{2}}=20000$ For wire $\mathrm{B}=\frac{l}{\mathrm{~d}^{2}}=\frac{100}{(0.1)^{2}}=10000$ For wire $\mathrm{C}=\frac{l}{\mathrm{~d}^{2}}=\frac{200}{(0.2)^{2}}=500$ For wire $\mathrm{D}=\frac{l}{\mathrm{~d}^{2}}=\frac{300}{(0.3)^{2}}=3333.33$ For highest resistance, the value of $\frac{l}{\mathrm{~d}^{2}}$ should be maximum.
UPSEE - 2004
Current Electricity
151946
Pick out the wrong statement from the following
1 The SI unit of conductance is mho
2 Conductance of a conductor decreases with increase in temperature
3 If the radius of a metallic wire is doubled, its resistance becomes $\left(\frac{1}{4}\right)^{\text {th }}$ of original resistance
4 If the length of the metallic wire is doubled, its resistivity remains uncharged
5 The relation between voltage and current for a non-ohmic conductor is linear
Explanation:
E (i) SI unit of the conductance is mho or siemens. (ii) The relation between resistance $\mathrm{R}$ and temperature $\mathrm{T}$ is as follows. $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)\right]$ Thus, as the resistance is increasing with increase in temperature, so the conductance will decrease with increase in temperature. (iii) The resistance is inversely proportional to square of radius of the wire. So, when the radius is doubled, the resistance will be $1 / 4^{\text {th }}$ of the original resistance. (iv) Resistivity is a constant term for a given material. It doesn't change by changing the length and crosssectional area of the material. (v) The voltage-current relation for ohmic conductor is linear but for a non-ohmic conductor it is non-linear.
Kerala CEE- 2013
Current Electricity
151949
The resistance of a $10 \mathrm{~m}$ long wire is $10 \Omega$. Its length is increased by $25 \%$ by stretching the wire uniformly. Then the resistance of the wire will be
151942
Two wires have lengths, diameters and specific resistance all in the ratio of $1: 2$. The resistance of the first wire is $10 \mathrm{ohm}$. Resistance of the second wire in ohm will be
151943
A light emitting diode (LED) has a voltage drop of $2 \mathrm{~V}$ across it and passes a current of $10 \mathrm{~mA}$. When it operates with a $6 \mathrm{~V}$ battery through a limiting resistor $R$, the value of $R$ is:
1 $40 \mathrm{k} \Omega$
2 $4 \mathrm{k} \Omega$
3 $200 \mathrm{k} \Omega$
4 $400 \Omega$
Explanation:
D Given, Current $(\mathrm{I})=10 \mathrm{~mA}$ $=0.01 \mathrm{~A}$ Voltage $\left(\mathrm{V}_{1}\right)=2 \mathrm{~V}$ Battery Voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Total Voltage $\left(\mathrm{V}_{\mathrm{T}}\right)=\mathrm{V}_{2}-\mathrm{V}_{1}=6-2$ By ohm's law, $=4 \mathrm{~V}$ $\mathrm{V}=\mathrm{IR}$ $4=0.01 \times R$ $\Rightarrow \quad \mathrm{R}=\frac{4}{0.01}=400 \Omega$
UPSEE - 2006
Current Electricity
151945
The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance?
A We know that, Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\frac{\pi d^{2}}{4}\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ For wire $\mathrm{A}=\frac{l}{\mathrm{~d}^{2}}=\frac{50}{(0.05)^{2}}=20000$ For wire $\mathrm{B}=\frac{l}{\mathrm{~d}^{2}}=\frac{100}{(0.1)^{2}}=10000$ For wire $\mathrm{C}=\frac{l}{\mathrm{~d}^{2}}=\frac{200}{(0.2)^{2}}=500$ For wire $\mathrm{D}=\frac{l}{\mathrm{~d}^{2}}=\frac{300}{(0.3)^{2}}=3333.33$ For highest resistance, the value of $\frac{l}{\mathrm{~d}^{2}}$ should be maximum.
UPSEE - 2004
Current Electricity
151946
Pick out the wrong statement from the following
1 The SI unit of conductance is mho
2 Conductance of a conductor decreases with increase in temperature
3 If the radius of a metallic wire is doubled, its resistance becomes $\left(\frac{1}{4}\right)^{\text {th }}$ of original resistance
4 If the length of the metallic wire is doubled, its resistivity remains uncharged
5 The relation between voltage and current for a non-ohmic conductor is linear
Explanation:
E (i) SI unit of the conductance is mho or siemens. (ii) The relation between resistance $\mathrm{R}$ and temperature $\mathrm{T}$ is as follows. $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)\right]$ Thus, as the resistance is increasing with increase in temperature, so the conductance will decrease with increase in temperature. (iii) The resistance is inversely proportional to square of radius of the wire. So, when the radius is doubled, the resistance will be $1 / 4^{\text {th }}$ of the original resistance. (iv) Resistivity is a constant term for a given material. It doesn't change by changing the length and crosssectional area of the material. (v) The voltage-current relation for ohmic conductor is linear but for a non-ohmic conductor it is non-linear.
Kerala CEE- 2013
Current Electricity
151949
The resistance of a $10 \mathrm{~m}$ long wire is $10 \Omega$. Its length is increased by $25 \%$ by stretching the wire uniformly. Then the resistance of the wire will be
151942
Two wires have lengths, diameters and specific resistance all in the ratio of $1: 2$. The resistance of the first wire is $10 \mathrm{ohm}$. Resistance of the second wire in ohm will be
151943
A light emitting diode (LED) has a voltage drop of $2 \mathrm{~V}$ across it and passes a current of $10 \mathrm{~mA}$. When it operates with a $6 \mathrm{~V}$ battery through a limiting resistor $R$, the value of $R$ is:
1 $40 \mathrm{k} \Omega$
2 $4 \mathrm{k} \Omega$
3 $200 \mathrm{k} \Omega$
4 $400 \Omega$
Explanation:
D Given, Current $(\mathrm{I})=10 \mathrm{~mA}$ $=0.01 \mathrm{~A}$ Voltage $\left(\mathrm{V}_{1}\right)=2 \mathrm{~V}$ Battery Voltage $\left(\mathrm{V}_{2}\right)=6 \mathrm{~V}$ Total Voltage $\left(\mathrm{V}_{\mathrm{T}}\right)=\mathrm{V}_{2}-\mathrm{V}_{1}=6-2$ By ohm's law, $=4 \mathrm{~V}$ $\mathrm{V}=\mathrm{IR}$ $4=0.01 \times R$ $\Rightarrow \quad \mathrm{R}=\frac{4}{0.01}=400 \Omega$
UPSEE - 2006
Current Electricity
151945
The following four wires are made of the same material and are at the same temperature. Which one of them has highest electrical resistance?
A We know that, Resistance $(\mathrm{R})=\frac{\rho l}{\mathrm{~A}}=\frac{\rho l}{\left(\frac{\pi d^{2}}{4}\right)}$ $\mathrm{R} \propto \frac{l}{\mathrm{~d}^{2}}$ For wire $\mathrm{A}=\frac{l}{\mathrm{~d}^{2}}=\frac{50}{(0.05)^{2}}=20000$ For wire $\mathrm{B}=\frac{l}{\mathrm{~d}^{2}}=\frac{100}{(0.1)^{2}}=10000$ For wire $\mathrm{C}=\frac{l}{\mathrm{~d}^{2}}=\frac{200}{(0.2)^{2}}=500$ For wire $\mathrm{D}=\frac{l}{\mathrm{~d}^{2}}=\frac{300}{(0.3)^{2}}=3333.33$ For highest resistance, the value of $\frac{l}{\mathrm{~d}^{2}}$ should be maximum.
UPSEE - 2004
Current Electricity
151946
Pick out the wrong statement from the following
1 The SI unit of conductance is mho
2 Conductance of a conductor decreases with increase in temperature
3 If the radius of a metallic wire is doubled, its resistance becomes $\left(\frac{1}{4}\right)^{\text {th }}$ of original resistance
4 If the length of the metallic wire is doubled, its resistivity remains uncharged
5 The relation between voltage and current for a non-ohmic conductor is linear
Explanation:
E (i) SI unit of the conductance is mho or siemens. (ii) The relation between resistance $\mathrm{R}$ and temperature $\mathrm{T}$ is as follows. $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{\mathrm{o}}\right)\right]$ Thus, as the resistance is increasing with increase in temperature, so the conductance will decrease with increase in temperature. (iii) The resistance is inversely proportional to square of radius of the wire. So, when the radius is doubled, the resistance will be $1 / 4^{\text {th }}$ of the original resistance. (iv) Resistivity is a constant term for a given material. It doesn't change by changing the length and crosssectional area of the material. (v) The voltage-current relation for ohmic conductor is linear but for a non-ohmic conductor it is non-linear.
Kerala CEE- 2013
Current Electricity
151949
The resistance of a $10 \mathrm{~m}$ long wire is $10 \Omega$. Its length is increased by $25 \%$ by stretching the wire uniformly. Then the resistance of the wire will be