266620
An infinite ladder is constructed with \(1 \$ \& 2 \$ 2\) resistors as shown in fig. Find the current that passes through the \(2 \Omega\) resistor nearest to the battery:
1 1.5 A
2 3 A
3 2 A
4 1 A
Explanation:
a Let \(x\) be the equvalent resistance between \(A\) and \(B\) which is also equal to between C and D . \[ \begin{aligned} & R_{e q}=x=\frac{2 x}{2+x}+1 \Rightarrow 2 x+x^2=2 x+2+x \\ & x^2-x-2=0 \\ & x^2-2 x+x-2=0 \\ & x(x-2)+1(x-2)=0 \\ & (x-2)(x+1)=0 \\ & \Rightarrow x=2 \& x=-1 \\ & \begin{array}{l} \therefore R_{e q}=2 \Omega \quad \therefore I=\frac{6}{2}=3 \mathrm{~A} \\ \quad \therefore I_1=I_2=1.5 \mathrm{~A} \end{array} \end{aligned} \]
**NCERT-XII-II-336**
TEST SERIES (PHYSICS FST)
266621
In the given figure potential difference between \(A\) and \(B\) is :
266622
The wavelength of radiation is \({ }_{\%}\) when an electron jumps from third to second orbit of hydrogen atom. For the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
266623
A proton and an electron both moving with the same velocity v enter into a region of uniform magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
1 Their time periods will be same
2 The time period for proton will be higher
3 The time period for electron will be higher
4 The orbital radii will be the same
Explanation:
b \(\mathrm{qvB}=m r \omega^2=\mathrm{mr} 4 \pi / \mathrm{T}^2\) or \(\mathrm{T} \propto \sqrt{\mathrm{m}} . \mathrm{As}\) mass of proton is higher than that of electron so \(\mathrm{T}_{\mathrm{p}}>\mathrm{T}_{\mathrm{e}}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
266620
An infinite ladder is constructed with \(1 \$ \& 2 \$ 2\) resistors as shown in fig. Find the current that passes through the \(2 \Omega\) resistor nearest to the battery:
1 1.5 A
2 3 A
3 2 A
4 1 A
Explanation:
a Let \(x\) be the equvalent resistance between \(A\) and \(B\) which is also equal to between C and D . \[ \begin{aligned} & R_{e q}=x=\frac{2 x}{2+x}+1 \Rightarrow 2 x+x^2=2 x+2+x \\ & x^2-x-2=0 \\ & x^2-2 x+x-2=0 \\ & x(x-2)+1(x-2)=0 \\ & (x-2)(x+1)=0 \\ & \Rightarrow x=2 \& x=-1 \\ & \begin{array}{l} \therefore R_{e q}=2 \Omega \quad \therefore I=\frac{6}{2}=3 \mathrm{~A} \\ \quad \therefore I_1=I_2=1.5 \mathrm{~A} \end{array} \end{aligned} \]
**NCERT-XII-II-336**
TEST SERIES (PHYSICS FST)
266621
In the given figure potential difference between \(A\) and \(B\) is :
266622
The wavelength of radiation is \({ }_{\%}\) when an electron jumps from third to second orbit of hydrogen atom. For the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
266623
A proton and an electron both moving with the same velocity v enter into a region of uniform magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
1 Their time periods will be same
2 The time period for proton will be higher
3 The time period for electron will be higher
4 The orbital radii will be the same
Explanation:
b \(\mathrm{qvB}=m r \omega^2=\mathrm{mr} 4 \pi / \mathrm{T}^2\) or \(\mathrm{T} \propto \sqrt{\mathrm{m}} . \mathrm{As}\) mass of proton is higher than that of electron so \(\mathrm{T}_{\mathrm{p}}>\mathrm{T}_{\mathrm{e}}\).
266620
An infinite ladder is constructed with \(1 \$ \& 2 \$ 2\) resistors as shown in fig. Find the current that passes through the \(2 \Omega\) resistor nearest to the battery:
1 1.5 A
2 3 A
3 2 A
4 1 A
Explanation:
a Let \(x\) be the equvalent resistance between \(A\) and \(B\) which is also equal to between C and D . \[ \begin{aligned} & R_{e q}=x=\frac{2 x}{2+x}+1 \Rightarrow 2 x+x^2=2 x+2+x \\ & x^2-x-2=0 \\ & x^2-2 x+x-2=0 \\ & x(x-2)+1(x-2)=0 \\ & (x-2)(x+1)=0 \\ & \Rightarrow x=2 \& x=-1 \\ & \begin{array}{l} \therefore R_{e q}=2 \Omega \quad \therefore I=\frac{6}{2}=3 \mathrm{~A} \\ \quad \therefore I_1=I_2=1.5 \mathrm{~A} \end{array} \end{aligned} \]
**NCERT-XII-II-336**
TEST SERIES (PHYSICS FST)
266621
In the given figure potential difference between \(A\) and \(B\) is :
266622
The wavelength of radiation is \({ }_{\%}\) when an electron jumps from third to second orbit of hydrogen atom. For the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
266623
A proton and an electron both moving with the same velocity v enter into a region of uniform magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
1 Their time periods will be same
2 The time period for proton will be higher
3 The time period for electron will be higher
4 The orbital radii will be the same
Explanation:
b \(\mathrm{qvB}=m r \omega^2=\mathrm{mr} 4 \pi / \mathrm{T}^2\) or \(\mathrm{T} \propto \sqrt{\mathrm{m}} . \mathrm{As}\) mass of proton is higher than that of electron so \(\mathrm{T}_{\mathrm{p}}>\mathrm{T}_{\mathrm{e}}\).
266620
An infinite ladder is constructed with \(1 \$ \& 2 \$ 2\) resistors as shown in fig. Find the current that passes through the \(2 \Omega\) resistor nearest to the battery:
1 1.5 A
2 3 A
3 2 A
4 1 A
Explanation:
a Let \(x\) be the equvalent resistance between \(A\) and \(B\) which is also equal to between C and D . \[ \begin{aligned} & R_{e q}=x=\frac{2 x}{2+x}+1 \Rightarrow 2 x+x^2=2 x+2+x \\ & x^2-x-2=0 \\ & x^2-2 x+x-2=0 \\ & x(x-2)+1(x-2)=0 \\ & (x-2)(x+1)=0 \\ & \Rightarrow x=2 \& x=-1 \\ & \begin{array}{l} \therefore R_{e q}=2 \Omega \quad \therefore I=\frac{6}{2}=3 \mathrm{~A} \\ \quad \therefore I_1=I_2=1.5 \mathrm{~A} \end{array} \end{aligned} \]
**NCERT-XII-II-336**
TEST SERIES (PHYSICS FST)
266621
In the given figure potential difference between \(A\) and \(B\) is :
266622
The wavelength of radiation is \({ }_{\%}\) when an electron jumps from third to second orbit of hydrogen atom. For the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be:
266623
A proton and an electron both moving with the same velocity v enter into a region of uniform magnetic field directed perpendicular to the velocity of the particles. They will now move in circular orbits such that
1 Their time periods will be same
2 The time period for proton will be higher
3 The time period for electron will be higher
4 The orbital radii will be the same
Explanation:
b \(\mathrm{qvB}=m r \omega^2=\mathrm{mr} 4 \pi / \mathrm{T}^2\) or \(\mathrm{T} \propto \sqrt{\mathrm{m}} . \mathrm{As}\) mass of proton is higher than that of electron so \(\mathrm{T}_{\mathrm{p}}>\mathrm{T}_{\mathrm{e}}\).
