Errors in Measurements
137 Questions
Start Allen Test
Q76
Allen
4. Semiconductor
MCQ
To get output '1' at R, for the given logic gate circuit the input values must be:
A.
$X = 0, Y = 1$
B.
$X = 1, Y = 1$
C.
$X = 0, Y = 0$
D.
$X = 1, Y = 0$
Q77
Allen
4. Semiconductor
MCQ
Boolean relation at the output stage-Y for the following circuit is:
A.
$A + B$
B.
$\vec{A}+\vec{B}$
C.
$\vec{A}\cdot\vec{B}$
D.
$A\cdot B$
Q78
Allen
1. Significant Digits
MSQ
A cuboid has sides of 10.4 cm, 3.8 cm and 1.02 cm. Choose the correct statement(s) about the area of the faces and its volume with due regard to significant digits.
A.
The largest face area is $39.52 \, cm^{2}$ .
B.
The smallest face area is $3.9 \, cm^{2}$ .
C.
The volume of the cuboid is $4.0 \times 10^{1} \, cm^{3}$ .
D.
The perimeter of the smallest face is 9.64 cm.
Q79
Allen
2. Vernier Callipers and Screw Gauge
MSQ
In ordinary Vernier callipers, 10 $^{th}$ division of the Vernier scale coincides with 9 $^{th}$ division of the main scale. In a specially designed Vernier callipers the Vernier scale is so constructed that 10 $^{th}$ division on it coincides with 11 $^{th}$ division on the main scale. Each division on the main scale equals to 1 mm. The callipers have a zero error as shown in the figure-I. When the Vernier calliper is used to measure a length, the concerned portion of its scale is shown in figure-II.
A.
Zero error in the callipers has magnitude 0.7 mm.
B.
The length being measured is 1.08 cm.
C.
The length being measured is 1.22 cm.
D.
Zero error in the callipers has magnitude 0.3 mm
Q80
Allen
2. Vernier Callipers and Screw Gauge
MSQ
want to design a carbon resistor in shape of a cylinder. Its resistivity is known precisely. The diameter is measured by a screw gauge whose main scale has a least count of 1 mm and circular scale has 100 divisions. The length is measured by a vernier callipers whose main scale is 1 mm and 9 main scale divisions coincide with 10 vernier scale divisions. wish that the fourth band in the colour code be painted gold. In which of the following screw gauge readings and vernier callipers readings will our goal be achieved?
A.
Screw gauge: msr 2 divisions, $50^{\text{th}}$ csd coinciding, Vernier scale: msr: $1.2 \, \text{cm}$ , $5^{\text{th}}$ vernier division coinciding.
B.
Screw gauge: msr 1 divisions, $0^{\text{th}}$ csd coinciding, Vernier scale: msr: $0.3 \, \text{cm}$ , $3^{\text{rd}}$ vernier division coinciding.
C.
Screw gauge: msr 0 divisions, $40^{\text{th}}$ csd coinciding, Vernier scale: msr: 5 cm, $0^{\text{th}}$ vernier division coinciding.
D.
Screw gauge: msr 1 divisions, $10^{\text{th}}$ csd coinciding, Vernier scale: msr: 8 cm, $8^{\text{th}}$ vernier division coinciding.
Q81
Allen
3. Error Analysis in Experiments
MSQ
In resonance air-column method, two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,
A.
the intensity of the sound heard at the first resonance was more than that at the second resonance.
B.
the prongs of the tuning fork were kept in a horizontal plane above the resonance tube.
C.
the amplitude of vibration of the ends of the prongs is typically around 1 cm.
D.
the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air.
Q82
Allen
3. Error Analysis in Experiments
MSQ
In an experiment to measure surface tension of completely wetting liquid, the capillary was dipped in liquid. The level of water in beaker was measured using travelling microscope to be 1.43cm. The level of water in capillary was measured to be 6.23cm. The readings for inner diameter were 0.00cm and 0.32cm respectively. Neglect the expansion of capillary and water on heating.
A.
% error in measurement of h is 0.4%
B.
% error in measurement of surface tension is 6.46%
C.
if temperature increases, level of water in capillary will be less than 6.23cm.
D.
$s = 0.38 \pm 0.04N / m$
Q83
Allen
3. Error Analysis in Experiments
MSQ
In the meter bridge circuit, the point D is the balance point.
A.
If the jockey is shifted to the left of D, current will flow from the meter bridge wire to the point B.
B.
If jockey is shifted to the right of $D$ , current will flow from point $B$ to the meter bridge wire.
C.
If the jockey is at point D, but now the resistance P is heated, then the current will flow from point B to the meter bridge wire.
D.
If the jockey is at point D, but now the resistance P is heated, then the current will flow from the meter bridge wire to the point B.
Q84
Allen
3. Error Analysis in Experiments
MSQ
In the Searle's experiment, after every step of loading, why should wait for two minutes before taking the readings?
A.
So that the wire can have its desired change in length.
B.
So that the wire can attain room temperature.
C.
So that vertical oscillations can get subsided.
D.
So that the wire has no change in its radius.
Q85
Allen
1. Significant Digits
Numerical
Significant figures are given in the following quantities are p, q, r, s, t, u, v and w respectively.
(A) 343 g (B) 2.20 (C) 1.103 N (D) 0.4142 s
(E) 0.0145 m (F) 1.0080 V (G) $9.1 \times 10^{4}$ km (H) $1.124 \times 10^{-3}$ V Value of $\frac{(p+q+r+s+t+u+v+w)}{4}$ is ____
(A) 343 g (B) 2.20 (C) 1.103 N (D) 0.4142 s
(E) 0.0145 m (F) 1.0080 V (G) $9.1 \times 10^{4}$ km (H) $1.124 \times 10^{-3}$ V Value of $\frac{(p+q+r+s+t+u+v+w)}{4}$ is ____
Correct Answer: 7
Question Type: Numerical
Explanation:
Ans. 7
$\begin{array}{l} \text { A n s . 7 } \\ \text { significant number are 3, 3, 4, 4, 3, 5, 2, and 4 of p, q, r, s, t, u, v and w Respectively } \\ \left(\frac {p + q + r + s + t + u + w}{4}\right) \\ = \left(\frac {3 + 3 + 4 + 4 + 3 + 5 + 2 + 4}{4}\right) \\ = \frac {2 8}{4} = 7 \end{array}$
$\begin{array}{l} \text { A n s . 7 } \\ \text { significant number are 3, 3, 4, 4, 3, 5, 2, and 4 of p, q, r, s, t, u, v and w Respectively } \\ \left(\frac {p + q + r + s + t + u + w}{4}\right) \\ = \left(\frac {3 + 3 + 4 + 4 + 3 + 5 + 2 + 4}{4}\right) \\ = \frac {2 8}{4} = 7 \end{array}$
Q86
Allen
2. Vernier Callipers and Screw Gauge
Numerical
In a vernier callipers, 19 divisions of its main scale match with (20) divisions on its vernier scale. Each division of the main scale is 'a' units. Using the vernier principle, if least count is $\frac{a}{4x}$ then find the value of $x$ ____.
Correct Answer: 5
Question Type: Numerical
Explanation:
Ans. 5
1 MSD = a
19 MSD = 20 VSD
ā 1VSD = $\frac{19}{20}$ MSD
ā 1VSD = $\frac{19}{20}$ a
L.C. = 1 MSD - 1 VSD
ā L.C. = a - $\frac{19a}{20}$ = $\frac{a}{20}$ ā L.C. = $\frac{a}{20}$ = $\frac{a}{4x}$ ā x = 5
1 MSD = a
19 MSD = 20 VSD
ā 1VSD = $\frac{19}{20}$ MSD
ā 1VSD = $\frac{19}{20}$ a
L.C. = 1 MSD - 1 VSD
ā L.C. = a - $\frac{19a}{20}$ = $\frac{a}{20}$ ā L.C. = $\frac{a}{20}$ = $\frac{a}{4x}$ ā x = 5
Q87
Allen
2. Vernier Callipers and Screw Gauge
Numerical
The pitch of a screw gauge is 1 mm and there are 50 divisions on its cap. When nothing is put in between the studs, 44 $^{th}$ division of the circular scale coincides with the reference line zero of the main scale is not visible. When a glass plate is placed between the studs, the main scale reads three divisions and the circular scale reads 26 divisions. Calculate the thickness of the plate.
Correct Answer: $R_{t} = 3.64 \mathrm{~mm}$
Question Type: Numerical
Explanation:
Ans. Rt = 3.64 mm
Pitch = 1 mm
Least count (LC) = $\frac{\text{Pitch}}{\text{Circular divisions}}$ ā L.C. = $\frac{1\text{mm}}{50}$ Zero error = - (50 - 44) Ć L.C.
ā Zero error = - 6 L.C.
Reading = MSR + VSR - zero error
= 3 mm + 26 Ć LC - (-6 LC)
= 3 mm + 32 LC
= 3 mm + $\frac{32\times 1\text{mm}}{50}$ Reading = 3mm + 0.64 mm = 3.64 mm
Pitch = 1 mm
Least count (LC) = $\frac{\text{Pitch}}{\text{Circular divisions}}$ ā L.C. = $\frac{1\text{mm}}{50}$ Zero error = - (50 - 44) Ć L.C.
ā Zero error = - 6 L.C.
Reading = MSR + VSR - zero error
= 3 mm + 26 Ć LC - (-6 LC)
= 3 mm + 32 LC
= 3 mm + $\frac{32\times 1\text{mm}}{50}$ Reading = 3mm + 0.64 mm = 3.64 mm
Q88
Allen
2. Vernier Callipers and Screw Gauge
Numerical
Using screw gauge, the observation of the diameter of a wire are 1.324, 1.326, 1.334, 1.336 cm respectively. Find the average diameter, the mean error, the relative error and % error.
Correct Answer: $\bar{D}=1.330cm.\overline{\Delta D}=0.005cm$ , Relative error = +0.004%, error = 0.4%
Question Type: Numerical
Explanation:
Ans. $\overline{D}$ = 1.330cm. $\overline{\Delta D}$ = 0.005cm, Relative error = + 0.004 %, error = 0.4% $D=\frac{\Sigma(D)}{N}=\frac{1.324+1.326+1.334+1.336}{4}$ = 1.330. $\Delta D_{1}$ = 1.324 - 1.330 = -0.006 $\Delta D_{2}$ = 1.326 - 1.330 = -0.004 $\Delta D_{3}$ = 1.334 - 1.330 = 0.004 $\Delta D_{4}$ = 1.336 - 1.330 = 0.006 $\Delta D=\frac{|\Delta D_{1}|+|\Delta D_{2}|+|\Delta D_{3}|+|\Delta D_{4}|}{4}$ = $\frac{0.006+0.004+0.004+0.006}{4}=\frac{0.020}{4}$ = 0.005cm
Relative error = $\frac{\Delta D}{D}=\frac{0.005}{1.330}$ = 0.004
% error = $\frac{\Delta D}{D}$ Ć 100 = 0.4%.
Relative error = $\frac{\Delta D}{D}=\frac{0.005}{1.330}$ = 0.004
% error = $\frac{\Delta D}{D}$ Ć 100 = 0.4%.
Q89
Allen
2. Vernier Callipers and Screw Gauge
Numerical
Consider a home made vernier scale as shown in the figure.
In this diagram, are interested in measuring the length of the line PQ. If both the inclines are identical and their angles are equal to $\theta$ then what is the least count of the instrument.
In this diagram, are interested in measuring the length of the line PQ. If both the inclines are identical and their angles are equal to $\theta$ then what is the least count of the instrument.
Correct Answer: $L.C. = \ell \left[\frac{1 - \cos\theta}{\cos\theta}\right]$
Question Type: Numerical
Explanation:
Ans. $L.C. = \ell \left[ \frac{1 - \cos \theta}{\cos \theta} \right]$ Least count $= \frac{\ell}{\cos \theta} - \ell$ $= \ell \left[ \frac{1}{\cos \theta} - 1 \right]$ $= \ell \left[ \frac{1 - \cos \theta}{\cos \theta} \right]$
Q90
Allen
3. Error Analysis in Experiments
Numerical
In a given optical bench, a needle of length 10 cm is used to estimate bench error. The object needle, image needle and lens holder have their reading as shown.
$x _ {0} = 1. 1 c m$
$x _ {I} = 2 1. 0 c m$
$x _ {L} = 1 0. 9 c m$
Estimate the bench errors which are present in image needle holder and object needle holder. Also find the focal length of the convex lens when.
$x _ {0} = 0. 6 c m$
$x _ {I} = 2 2. 5 c m$
$x _ {L} = 1 1. 4 c m$
Correct Answer: $5.5\pm 0.1cm$Question Type: Numerical
Explanation:
Ans. 5.5 Ā± 0.1 cm
Bench error in measurement of object distance from lens. $u_{\text{bench error}} = (x_L - x_0) - 10 \, \text{cm}$ $u_{\text{bench error}} = 9.8 - 10 = -0.2 \, \text{cm}$ Bench error in measurement of image distance from lens. $V_{\text{bench error}} = (x_I - x_L) - 10 \, \text{cm}$ $V_{\text{bench error}} = (10.1) - 10 \, \text{cm} = +0.1 \, \text{cm}$ Now focal length for given reading $x_0 = 0.6 \, \text{cm} \quad x_I = 22.5 \, \text{cm} \quad x_L = 11.4 \, \text{cm}$ $u = -[(x_L - x_0) - \text{bench error}]$ $u = -[(11.4 - 0.6)cm - (-0.2)cm]$ $u = -11 \, \text{cm}$ $V = (x_I - x_L) - \text{bench error}$ $V = 11.1 \, \text{cm} - 0.1 \, \text{cm} = 11 \, \text{cm}$ $\frac{1}{f} = \frac{1}{V} - \frac{1}{u}$ $\Rightarrow \frac{1}{f} = \frac{1}{11} - \frac{1}{(-11)}$ $\Rightarrow f = \frac{11}{2} cm = 5.5 \, cm$ $\frac{\Delta f}{f^2} = \frac{\Delta V}{V^2} + \frac{\Delta u}{u^2}$ $\Rightarrow \Delta f = \left( \frac{\Delta V}{V^2} + \frac{\Delta u}{u^2} \right) f^2$ $\Rightarrow u = x_L - x_0, V = x_I - x_L$ $\Delta u = \Delta x_L + \Delta x_0$ $\Delta u = 0.1 + 0.1 = 0.2$ $\Delta f = \left( \frac{0.2}{11^2} + \frac{0.2}{11^2} \right) \times \left( \frac{11}{2} \right)^2$ $\Delta f = \frac{0.4}{11^2} \times \frac{11^2}{4} = 0.1$ $f = 5.5 \pm 0.1 \, cm$
Bench error in measurement of object distance from lens. $u_{\text{bench error}} = (x_L - x_0) - 10 \, \text{cm}$ $u_{\text{bench error}} = 9.8 - 10 = -0.2 \, \text{cm}$ Bench error in measurement of image distance from lens. $V_{\text{bench error}} = (x_I - x_L) - 10 \, \text{cm}$ $V_{\text{bench error}} = (10.1) - 10 \, \text{cm} = +0.1 \, \text{cm}$ Now focal length for given reading $x_0 = 0.6 \, \text{cm} \quad x_I = 22.5 \, \text{cm} \quad x_L = 11.4 \, \text{cm}$ $u = -[(x_L - x_0) - \text{bench error}]$ $u = -[(11.4 - 0.6)cm - (-0.2)cm]$ $u = -11 \, \text{cm}$ $V = (x_I - x_L) - \text{bench error}$ $V = 11.1 \, \text{cm} - 0.1 \, \text{cm} = 11 \, \text{cm}$ $\frac{1}{f} = \frac{1}{V} - \frac{1}{u}$ $\Rightarrow \frac{1}{f} = \frac{1}{11} - \frac{1}{(-11)}$ $\Rightarrow f = \frac{11}{2} cm = 5.5 \, cm$ $\frac{\Delta f}{f^2} = \frac{\Delta V}{V^2} + \frac{\Delta u}{u^2}$ $\Rightarrow \Delta f = \left( \frac{\Delta V}{V^2} + \frac{\Delta u}{u^2} \right) f^2$ $\Rightarrow u = x_L - x_0, V = x_I - x_L$ $\Delta u = \Delta x_L + \Delta x_0$ $\Delta u = 0.1 + 0.1 = 0.2$ $\Delta f = \left( \frac{0.2}{11^2} + \frac{0.2}{11^2} \right) \times \left( \frac{11}{2} \right)^2$ $\Delta f = \frac{0.4}{11^2} \times \frac{11^2}{4} = 0.1$ $f = 5.5 \pm 0.1 \, cm$
Q91
Allen
3. Error Analysis in Experiments
Numerical
The energy of a system as a function of time t is given as $E(t) = A^{2}\exp(-\alpha t)$ , where $\alpha = 0.2 s^{-1}$ . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E(t)$ at t = 5 s is
Correct Answer: 4
Question Type: Numerical
Explanation:
Ans. 4 $E(t) = A^{2} e^{-\alpha t}$ $\ln E = \ln A^{2} + \ln e^{-\alpha t}$ $\ln E = 2 \ln a - \alpha t$ $\frac{\Delta E}{E} \times 100 = 2 \frac{\Delta A}{A} \times 100 + \alpha \Delta t \times 100$ $= 2 \times 1.25 + \frac{2}{10} \times 1.5 \times 5$ $= 2.5 + 1.5 = 4\%$
Q92
Allen
3. Error Analysis in Experiments
Numerical
A travelling microscope is used to measure the refractive index of a glass slab. The microscope is focused on a spot on the table and the reading of microscope is 10.08 cm. Now the glass slab is put on the spot and microscope is raised to focus on the apparent position of spot. The reading is 12.48 cm. Now the microscope is raised further to focus on the top of the slab. The reading is 14.88 cm. If the least count of scale is 0.01 cm, what is the % error in refractive index?
Correct Answer: 1.25
Question Type: Numerical
Explanation:
Ans. 1.25
$\frac {d _ {I}}{i} = \frac {d _ {0}}{\mu} = \frac {h _ {3} - h _ {1}}{\mu}$
$d _ {I} = h _ {3} - h _ {2}$
$\Rightarrow \mu = \frac {h _ {3} - h _ {1}}{h _ {3} - h _ {2}} = \frac {4 . 8 0}{2 . 4 0} = 2$
$\frac {d u}{\mu} = \frac {d h _ {3} + d h _ {1}}{h _ {3} - h _ {1}} + \frac {d h _ {3} + d h _ {2}}{h _ {3} - h _ {2}}$
$= \frac {0 . 0 2 [ 3 ]}{4 . 8 0} \times 1 0 0$
$\% \text{error} = \frac {60}{48}$
$\frac {d _ {I}}{i} = \frac {d _ {0}}{\mu} = \frac {h _ {3} - h _ {1}}{\mu}$
$d _ {I} = h _ {3} - h _ {2}$
$\Rightarrow \mu = \frac {h _ {3} - h _ {1}}{h _ {3} - h _ {2}} = \frac {4 . 8 0}{2 . 4 0} = 2$
$\frac {d u}{\mu} = \frac {d h _ {3} + d h _ {1}}{h _ {3} - h _ {1}} + \frac {d h _ {3} + d h _ {2}}{h _ {3} - h _ {2}}$
$= \frac {0 . 0 2 [ 3 ]}{4 . 8 0} \times 1 0 0$
$\% \text{error} = \frac {60}{48}$
Q93
Allen
4. Semiconductor
Numerical
For the circuit shown in figure, find:
(I) the output voltage
(II) the voltage drop across series resistance
(III) the current through zener diode
(I) the output voltage
(II) the voltage drop across series resistance
(III) the current through zener diode
Correct Answer: (i) $10V$ (ii) $90V$ (iii) $4mA$
Question Type: Numerical
Explanation:
Ans. (i) 10 V (ii) 90 V (iii) 4 mA
(i) Here output voltage will be equal to zener voltage
(i) Here output volta $V_{L}=10V$ (ii) $V_{R}=V_{total}-10V$ = 100 - 10 $V_{R}=90V$ (iii) $I_{L}=\frac{10}{5\times10^{3}}$ $=2\times10^{-3}A$ $I_{L}=2mA$ $I_{t}=\frac{90}{15\times10^{3}}A$ $=6\times10^{-3}A$ = 6 mA
By KCL $I_{t}=I_{Z}+I_{L}$ 6 mA = $I_{Z} + 2mA$ $I_{Z}=4mA$
(i) Here output voltage will be equal to zener voltage
(i) Here output volta $V_{L}=10V$ (ii) $V_{R}=V_{total}-10V$ = 100 - 10 $V_{R}=90V$ (iii) $I_{L}=\frac{10}{5\times10^{3}}$ $=2\times10^{-3}A$ $I_{L}=2mA$ $I_{t}=\frac{90}{15\times10^{3}}A$ $=6\times10^{-3}A$ = 6 mA
By KCL $I_{t}=I_{Z}+I_{L}$ 6 mA = $I_{Z} + 2mA$ $I_{Z}=4mA$
Q94
Allen
4. Semiconductor
Numerical
Write a truth table for the circuit in figure, including the states at C, D, E, F and G.
Correct Answer: A B C D E F G
0 0 1 1 0 0 1
0 1 1 0 0 1 0
1 0 0 1 1 0 0
1 1 0 0 0 0 1
Question Type: Numerical
Explanation:
Ans. (54)
A B C D E F G
0 0 1 1 0 0 1
0 1 1 0 0 1 0
1 0 0 1 1 0 0
1 1 0 0 0 0 1
$\begin{array}{c c c c c c c} \text {A} & \text {B} & \text {C} & \text {D} & \text {E} & \text {F} & \text {G} \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 \end{array}$
A B C D E F G
0 0 1 1 0 0 1
0 1 1 0 0 1 0
1 0 0 1 1 0 0
1 1 0 0 0 0 1
$\begin{array}{c c c c c c c} \text {A} & \text {B} & \text {C} & \text {D} & \text {E} & \text {F} & \text {G} \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 \end{array}$
Q95
Allen
6. JEE Main Practice Paper
MCQ
If $a = 8 \pm 0.08$ and $b = 6 \pm 0.06$ . Let $x = a + b$ , $y = a - b$ , $z = a \times b$ . The correct order of % error in x, y and z is :-
A.
x = y < z
B.
x = y > z
C.
x < z < y
D.
x > z < y
Q96
Allen
6. JEE Main Practice Paper
MCQ
The edge of a cube is $a = 1.2 \times 10^{-2} m$ . Then its volume will be recorded as:
A.
$1.7 \times 10^{-6} m^{3}$
B.
$1.70 \times 10^{-6} m^{3}$
C.
$1.70 \times 10^{-7} m^{3}$
D.
$1.78 \times 10^{-6} m^{3}$
Q97
Allen
6. JEE Main Practice Paper
MCQ
On the basis of detail given about two measuring instruments, select the correct statement.
(i) Vernier callipers having main scale division = 0.05 cm and Vernier scale division = $\frac{2.45}{50}$ cm.
(ii) Screw gauge having pitch 0.5 mm and its circular scale division measures 0.01 mm.
A.
Both the instruments have same least count.
B.
Least count of screw gauge is lesser than that of vernier callipers.
C.
Both the instruments have same least count but screw gauge is more precise.
D.
Both the instruments have different least count and screw gauge is more precise.
Q98
Allen
6. JEE Main Practice Paper
MCQ
In a Searle's experiment for determination of Young's Modulus, when a load of 50 kg is added to a 3 meter long wire micrometer screw having pitch 1 mm needs to be given a quarter turn in order to restore the horizontal position of spirit level. Young's modulus of the wire if its cross-sectional area is $10^{-5}$ m $^{2}$ is
A.
$6 \times 10^{11} N/m^{2}$
B.
$1.5 \times 10^{11} N/m^{2}$
C.
$3 \times 10^{11} N/m^{2}$
D.
None
Q99
Allen
6. JEE Main Practice Paper
MCQ
An electric field is applied to a semi-conductor. Let the number of charge carriers be n and the average drift speed be v. If the temperature is increased:
A.
Both n and v will increase
B.
n will increase but v will decrease
C.
v will increase but n will decrease
D.
Both n and v will decrease
Q100
Allen
6. JEE Main Practice Paper
MCQ
P-type semiconductor is formed when
A. As impurity is mixed in Si
B. Al impurity is mixed in Si
C. B impurity is mixed in Ge
D. P impurity is mixed in Ge
A.
A and C
B.
A and D
C.
B and C
D.
B and D