263838
A coil of inductive reactance \(1 / \sqrt{3} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) supply. The time lag between maximum voltage and current is:
263839
In the circuit, the forward resistance of each diode is \(50 \Omega\) and the reverse resistance of each diode is infinite. Find the current through 20 ohm resistor:
1 0.2 A
2 0.4 A
3 0.3 A
4 0.1 A .
Explanation:
d Since \(\mathrm{D}_2\) is reverse biased, it will not conduct and \(D_1\) is forward biased, so its resistance will be \(50 \Omega\). \(\therefore \quad\) Current \(\mathrm{I}=\frac{10}{50+30+20}=0.1 \mathrm{~A}\)
NCERT-XII-1-33i
TEST SERIES (PHYSICS FST)
263840
A mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size. Calculate the energy expanded if surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nfm}\) :
1 \(0.156 \times 10^{-3} \mathrm{~J}\)
2 \(3.356 \times 10^{-3} \mathrm{~J}\)
3 \(4.356 \times 10^{-3} \mathrm{~J}\)
4 \(3.560 \times 10^{-3} \mathrm{~J}\).
Explanation:
c Initial surface area \(=4 \pi \mathrm{R}^2\) Final surface area \(=n\left(4 \pi r^2\right)\) Increase in area \(\Delta S=n\left(4 \pi r^2\right)-4 \pi R^2\) So, energy expanded in the process \(W=T \Delta A=4 \pi T\left(n r^2-R^2\right)\) Now since the total wolume of \(n\) droplets is the same as that of initial drop, i.e., \[ \frac{4}{3} \pi \mathrm{R}^3=\pi\left[\frac{4}{3} \pi \pi^3\right] \text { or } \mathrm{r}=\mathrm{R} / \pi^{1 / 3} \] So, by substituting the value of r from eqn (ii) in (i) \[ W=4 \pi R^2 \mathrm{~T}\left[(\pi)^{1 / 3}-1\right] \] So, here \[ \begin{aligned} & W=4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \times 35 \times 10^{-3}\left[10^2-1\right] \\ & =4.356 \times 10^{-3} \mathrm{~J} \end{aligned} \]
NCERT-XI-II-193
TEST SERIES (PHYSICS FST)
263841
Two open pipes \(A\) and \(B\) are sounded together such that 2 beats are heard. Now \(A\) is closed beat frequency is still 2. What is the ratio of length's of pipe:
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 1\)
4 \(4: 1\)
Explanation:
b \[ \frac{L_1}{L_2}=\frac{3}{4} \]
NCERT-XI-II-292
TEST SERIES (PHYSICS FST)
263842
Sodium and copper havework function 2.3 eV and 4.5 eV res pectively then the ratio of their threshold wavelengths is nearest to:
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
c \[ \frac{h c}{\lambda_0}=W_0 \Rightarrow \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(W_0\right)_2}{\left(W_0\right)_1}=\frac{4.5}{2.3}=\frac{2}{1} \]
263838
A coil of inductive reactance \(1 / \sqrt{3} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) supply. The time lag between maximum voltage and current is:
263839
In the circuit, the forward resistance of each diode is \(50 \Omega\) and the reverse resistance of each diode is infinite. Find the current through 20 ohm resistor:
1 0.2 A
2 0.4 A
3 0.3 A
4 0.1 A .
Explanation:
d Since \(\mathrm{D}_2\) is reverse biased, it will not conduct and \(D_1\) is forward biased, so its resistance will be \(50 \Omega\). \(\therefore \quad\) Current \(\mathrm{I}=\frac{10}{50+30+20}=0.1 \mathrm{~A}\)
NCERT-XII-1-33i
TEST SERIES (PHYSICS FST)
263840
A mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size. Calculate the energy expanded if surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nfm}\) :
1 \(0.156 \times 10^{-3} \mathrm{~J}\)
2 \(3.356 \times 10^{-3} \mathrm{~J}\)
3 \(4.356 \times 10^{-3} \mathrm{~J}\)
4 \(3.560 \times 10^{-3} \mathrm{~J}\).
Explanation:
c Initial surface area \(=4 \pi \mathrm{R}^2\) Final surface area \(=n\left(4 \pi r^2\right)\) Increase in area \(\Delta S=n\left(4 \pi r^2\right)-4 \pi R^2\) So, energy expanded in the process \(W=T \Delta A=4 \pi T\left(n r^2-R^2\right)\) Now since the total wolume of \(n\) droplets is the same as that of initial drop, i.e., \[ \frac{4}{3} \pi \mathrm{R}^3=\pi\left[\frac{4}{3} \pi \pi^3\right] \text { or } \mathrm{r}=\mathrm{R} / \pi^{1 / 3} \] So, by substituting the value of r from eqn (ii) in (i) \[ W=4 \pi R^2 \mathrm{~T}\left[(\pi)^{1 / 3}-1\right] \] So, here \[ \begin{aligned} & W=4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \times 35 \times 10^{-3}\left[10^2-1\right] \\ & =4.356 \times 10^{-3} \mathrm{~J} \end{aligned} \]
NCERT-XI-II-193
TEST SERIES (PHYSICS FST)
263841
Two open pipes \(A\) and \(B\) are sounded together such that 2 beats are heard. Now \(A\) is closed beat frequency is still 2. What is the ratio of length's of pipe:
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 1\)
4 \(4: 1\)
Explanation:
b \[ \frac{L_1}{L_2}=\frac{3}{4} \]
NCERT-XI-II-292
TEST SERIES (PHYSICS FST)
263842
Sodium and copper havework function 2.3 eV and 4.5 eV res pectively then the ratio of their threshold wavelengths is nearest to:
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
c \[ \frac{h c}{\lambda_0}=W_0 \Rightarrow \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(W_0\right)_2}{\left(W_0\right)_1}=\frac{4.5}{2.3}=\frac{2}{1} \]
263838
A coil of inductive reactance \(1 / \sqrt{3} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) supply. The time lag between maximum voltage and current is:
263839
In the circuit, the forward resistance of each diode is \(50 \Omega\) and the reverse resistance of each diode is infinite. Find the current through 20 ohm resistor:
1 0.2 A
2 0.4 A
3 0.3 A
4 0.1 A .
Explanation:
d Since \(\mathrm{D}_2\) is reverse biased, it will not conduct and \(D_1\) is forward biased, so its resistance will be \(50 \Omega\). \(\therefore \quad\) Current \(\mathrm{I}=\frac{10}{50+30+20}=0.1 \mathrm{~A}\)
NCERT-XII-1-33i
TEST SERIES (PHYSICS FST)
263840
A mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size. Calculate the energy expanded if surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nfm}\) :
1 \(0.156 \times 10^{-3} \mathrm{~J}\)
2 \(3.356 \times 10^{-3} \mathrm{~J}\)
3 \(4.356 \times 10^{-3} \mathrm{~J}\)
4 \(3.560 \times 10^{-3} \mathrm{~J}\).
Explanation:
c Initial surface area \(=4 \pi \mathrm{R}^2\) Final surface area \(=n\left(4 \pi r^2\right)\) Increase in area \(\Delta S=n\left(4 \pi r^2\right)-4 \pi R^2\) So, energy expanded in the process \(W=T \Delta A=4 \pi T\left(n r^2-R^2\right)\) Now since the total wolume of \(n\) droplets is the same as that of initial drop, i.e., \[ \frac{4}{3} \pi \mathrm{R}^3=\pi\left[\frac{4}{3} \pi \pi^3\right] \text { or } \mathrm{r}=\mathrm{R} / \pi^{1 / 3} \] So, by substituting the value of r from eqn (ii) in (i) \[ W=4 \pi R^2 \mathrm{~T}\left[(\pi)^{1 / 3}-1\right] \] So, here \[ \begin{aligned} & W=4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \times 35 \times 10^{-3}\left[10^2-1\right] \\ & =4.356 \times 10^{-3} \mathrm{~J} \end{aligned} \]
NCERT-XI-II-193
TEST SERIES (PHYSICS FST)
263841
Two open pipes \(A\) and \(B\) are sounded together such that 2 beats are heard. Now \(A\) is closed beat frequency is still 2. What is the ratio of length's of pipe:
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 1\)
4 \(4: 1\)
Explanation:
b \[ \frac{L_1}{L_2}=\frac{3}{4} \]
NCERT-XI-II-292
TEST SERIES (PHYSICS FST)
263842
Sodium and copper havework function 2.3 eV and 4.5 eV res pectively then the ratio of their threshold wavelengths is nearest to:
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
c \[ \frac{h c}{\lambda_0}=W_0 \Rightarrow \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(W_0\right)_2}{\left(W_0\right)_1}=\frac{4.5}{2.3}=\frac{2}{1} \]
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
TEST SERIES (PHYSICS FST)
263838
A coil of inductive reactance \(1 / \sqrt{3} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) supply. The time lag between maximum voltage and current is:
263839
In the circuit, the forward resistance of each diode is \(50 \Omega\) and the reverse resistance of each diode is infinite. Find the current through 20 ohm resistor:
1 0.2 A
2 0.4 A
3 0.3 A
4 0.1 A .
Explanation:
d Since \(\mathrm{D}_2\) is reverse biased, it will not conduct and \(D_1\) is forward biased, so its resistance will be \(50 \Omega\). \(\therefore \quad\) Current \(\mathrm{I}=\frac{10}{50+30+20}=0.1 \mathrm{~A}\)
NCERT-XII-1-33i
TEST SERIES (PHYSICS FST)
263840
A mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size. Calculate the energy expanded if surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nfm}\) :
1 \(0.156 \times 10^{-3} \mathrm{~J}\)
2 \(3.356 \times 10^{-3} \mathrm{~J}\)
3 \(4.356 \times 10^{-3} \mathrm{~J}\)
4 \(3.560 \times 10^{-3} \mathrm{~J}\).
Explanation:
c Initial surface area \(=4 \pi \mathrm{R}^2\) Final surface area \(=n\left(4 \pi r^2\right)\) Increase in area \(\Delta S=n\left(4 \pi r^2\right)-4 \pi R^2\) So, energy expanded in the process \(W=T \Delta A=4 \pi T\left(n r^2-R^2\right)\) Now since the total wolume of \(n\) droplets is the same as that of initial drop, i.e., \[ \frac{4}{3} \pi \mathrm{R}^3=\pi\left[\frac{4}{3} \pi \pi^3\right] \text { or } \mathrm{r}=\mathrm{R} / \pi^{1 / 3} \] So, by substituting the value of r from eqn (ii) in (i) \[ W=4 \pi R^2 \mathrm{~T}\left[(\pi)^{1 / 3}-1\right] \] So, here \[ \begin{aligned} & W=4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \times 35 \times 10^{-3}\left[10^2-1\right] \\ & =4.356 \times 10^{-3} \mathrm{~J} \end{aligned} \]
NCERT-XI-II-193
TEST SERIES (PHYSICS FST)
263841
Two open pipes \(A\) and \(B\) are sounded together such that 2 beats are heard. Now \(A\) is closed beat frequency is still 2. What is the ratio of length's of pipe:
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 1\)
4 \(4: 1\)
Explanation:
b \[ \frac{L_1}{L_2}=\frac{3}{4} \]
NCERT-XI-II-292
TEST SERIES (PHYSICS FST)
263842
Sodium and copper havework function 2.3 eV and 4.5 eV res pectively then the ratio of their threshold wavelengths is nearest to:
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
c \[ \frac{h c}{\lambda_0}=W_0 \Rightarrow \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(W_0\right)_2}{\left(W_0\right)_1}=\frac{4.5}{2.3}=\frac{2}{1} \]
263838
A coil of inductive reactance \(1 / \sqrt{3} \Omega\) and resistance \(1 \Omega\) is connected to a \(200 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) supply. The time lag between maximum voltage and current is:
263839
In the circuit, the forward resistance of each diode is \(50 \Omega\) and the reverse resistance of each diode is infinite. Find the current through 20 ohm resistor:
1 0.2 A
2 0.4 A
3 0.3 A
4 0.1 A .
Explanation:
d Since \(\mathrm{D}_2\) is reverse biased, it will not conduct and \(D_1\) is forward biased, so its resistance will be \(50 \Omega\). \(\therefore \quad\) Current \(\mathrm{I}=\frac{10}{50+30+20}=0.1 \mathrm{~A}\)
NCERT-XII-1-33i
TEST SERIES (PHYSICS FST)
263840
A mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size. Calculate the energy expanded if surface tension of mercury is \(35 \times 10^{-3} \mathrm{Nfm}\) :
1 \(0.156 \times 10^{-3} \mathrm{~J}\)
2 \(3.356 \times 10^{-3} \mathrm{~J}\)
3 \(4.356 \times 10^{-3} \mathrm{~J}\)
4 \(3.560 \times 10^{-3} \mathrm{~J}\).
Explanation:
c Initial surface area \(=4 \pi \mathrm{R}^2\) Final surface area \(=n\left(4 \pi r^2\right)\) Increase in area \(\Delta S=n\left(4 \pi r^2\right)-4 \pi R^2\) So, energy expanded in the process \(W=T \Delta A=4 \pi T\left(n r^2-R^2\right)\) Now since the total wolume of \(n\) droplets is the same as that of initial drop, i.e., \[ \frac{4}{3} \pi \mathrm{R}^3=\pi\left[\frac{4}{3} \pi \pi^3\right] \text { or } \mathrm{r}=\mathrm{R} / \pi^{1 / 3} \] So, by substituting the value of r from eqn (ii) in (i) \[ W=4 \pi R^2 \mathrm{~T}\left[(\pi)^{1 / 3}-1\right] \] So, here \[ \begin{aligned} & W=4 \times 3.14 \times\left(1 \times 10^{-2}\right)^2 \times 35 \times 10^{-3}\left[10^2-1\right] \\ & =4.356 \times 10^{-3} \mathrm{~J} \end{aligned} \]
NCERT-XI-II-193
TEST SERIES (PHYSICS FST)
263841
Two open pipes \(A\) and \(B\) are sounded together such that 2 beats are heard. Now \(A\) is closed beat frequency is still 2. What is the ratio of length's of pipe:
1 \(1: 1\)
2 \(3: 4\)
3 \(4: 1\)
4 \(4: 1\)
Explanation:
b \[ \frac{L_1}{L_2}=\frac{3}{4} \]
NCERT-XI-II-292
TEST SERIES (PHYSICS FST)
263842
Sodium and copper havework function 2.3 eV and 4.5 eV res pectively then the ratio of their threshold wavelengths is nearest to:
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
c \[ \frac{h c}{\lambda_0}=W_0 \Rightarrow \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(W_0\right)_2}{\left(W_0\right)_1}=\frac{4.5}{2.3}=\frac{2}{1} \]
