Errors in Measurements
137 Questions
Start Allen Test
Q101
Allen
6. JEE Main Practice Paper
MCQ
A strip of copper and another of germanium are cooled from room temperature to 80K. The resistance of -
A.
copper strip increases and that of germanium decreases
B.
copper strip decreases and that of germanium increases
C.
each of these increases
D.
each of these decreases
Q102
Allen
6. JEE Main Practice Paper
MCQ
The V-I characteristic for a p-n junction diode is plotted as shown in the figure. From the plot can conclude that
$[V_b \to$ breakdown voltage, $V_k \to$ knee voltage]
$[V_b \to$ breakdown voltage, $V_k \to$ knee voltage]
A.
The forward bias resistance of diode is very high; almost infinity for small values of $V$ and after a certain value it becomes very low.
B.
The reverse bias resistance of diode is very high in the beginning up to breakdown voltage is not achieved.
C.
Both forward and reverse bias resistances are same for all voltages.
D.
Both (1) and (2) are correct.
Q103
Allen
6. JEE Main Practice Paper
MCQ
A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery's positive terminal is connected to A and in other case when positive terminal of battery is connected to B will respectively be :-
A.
$0.1A$ and $0.2A$
B.
$0.4A$ and $0.2A$
C.
$0.2A$ and $0.4A$
D.
$0.2A$ and $0.1A$
PSC005
PSC005
Q104
Allen
6. JEE Main Practice Paper
MCQ
In the half-wave rectifier circuit shown. Which one of the following wave forms is true for VCD, if the input is as shown?
A.
B.
C.
D.
Q105
Allen
6. JEE Main Practice Paper
MCQ
Name the logic gate equivalent to the diagram attached
A.
OR
B.
NOR
C.
NAND
D.
AND
Q106
Allen
6. JEE Main Practice Paper
MCQ
The logic performed by the circuit shown in figure is equivalent to :
A.
AND
B.
NAND
C.
OR
D.
NOR
Q107
Allen
6. JEE Main Practice Paper
MCQ
The output from a NAND gate having inputs A and B given below will be.
A.
B.
C.
D.
Q108
Allen
6. JEE Main Practice Paper
MCQ
Given below are four logic gate symbols (figure). Those for OR, NOR and NAND are respectively.
A.
1, 4, 3
B.
4, 1, 2
C.
1, 3, 4
D.
4, 2, 1
Q109
Allen
6. JEE Main Practice Paper
MCQ
The following truth table corresponds to the logic gate -
A- 0 0 1 1
B- 0 1 0 1
X- 0 1 1 1
A.
NAND
B.
OR
C.
AND
D.
XOR
Q110
Allen
6. JEE Main Practice Paper
MCQ
The following figure shows a logic gate circuit with two input A and B output C. The voltage waveforms of A, B and C are as shown in second figure below. The logic gate is:
A.
OR gate
B.
NAND gate
C.
AND gate
D.
NOR gate
Q111
Allen
6. JEE Main Practice Paper
MCQ
The combination of 'NAND' gates shown here under (figure) are equivalent to -
A.
An OR gate and an AND gate respectively
B.
An AND gate and a NOT gate respectively
C.
An AND gate and an OR gate respectively
D.
An OR gate and a NOT gate respectively
Q112
Allen
6. JEE Main Practice Paper
Numerical
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $\mu$ m diameter of wire is:
Correct Answer: 200
Explanation:
Ans. (200) $L.C. = \frac{pitch}{no. of divisions}$ $5 \times 10^{-6} m = \frac{1mm}{n} = \frac{10^{-3}m}{n}$ $\Rightarrow n = \frac{1000}{5} = 200$
Q113
Allen
6. JEE Main Practice Paper
Numerical
In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with ninth main scale division. Then the value of vernier constant will be $\cdots\cdots\times10^{-2}$ mm.
Correct Answer: 5
Explanation:
Ans. (5) $20 MSD = 1cm$ $1MSD = \frac{1}{20} cm$ $10 VSD = 9MSD$ $1VSD = \frac{9}{10} MSD$ $= \frac{9}{10} \times \frac{1}{20} cm$ $1VSD = \frac{9}{200} cm$ $VC = 1 MSD - 1 VSD$ $= \frac{1}{20} cm - \frac{9}{200} cm$ $= \frac{1}{200} \times 10mm$ $VC = 5 \times 10^{-2} mm$
Q114
Allen
6. JEE Main Practice Paper
Numerical
Student A and Student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ____. [Figure shows position of reference 'O' when jaws of screw gauge are closed] Given pitch = 0.1 cm.
Correct Answer: 13
Explanation:
Ans. (13)
For (A)
Reading = $MSR + CSR + Error$ $0.322 = 0.300 + CSR + 5 \times LC$ $0.322 = 0.300 + CSR + 0.005$ $CSR = 0.017$ For (B)
Reading = $MSR + CSR + Error$ $0.322 = 0.200 + CSR + 0.092$ $CSR = 0.030$ Difference = $0.030 - 0.017 = 0.013 cm$ Division on circular scale = $\frac{0.013}{0.001} = 13$
For (A)
Reading = $MSR + CSR + Error$ $0.322 = 0.300 + CSR + 5 \times LC$ $0.322 = 0.300 + CSR + 0.005$ $CSR = 0.017$ For (B)
Reading = $MSR + CSR + Error$ $0.322 = 0.200 + CSR + 0.092$ $CSR = 0.030$ Difference = $0.030 - 0.017 = 0.013 cm$ Division on circular scale = $\frac{0.013}{0.001} = 13$
Q115
Allen
6. JEE Main Practice Paper
Numerical
The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R (in Ω) connected in series with the diode for obtaining maximum current –
Correct Answer: 5
Explanation:
Ans. (5) $VI = 100 \times 10^{-3}W \Rightarrow I = \frac{10^{-1}}{V} = \frac{10^{-1}}{0.5} = \frac{1}{5}A$ Applying KVL: $1.5 - \frac{1}{5}R - 0.5 = 0 \Rightarrow R = 5\Omega$
Q116
Allen
6. JEE Main Practice Paper
Numerical
A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the circuit shown so that half wave rectification occurs. If the forward resistance of the diode is negligible compared to R then rms voltage (in volt) across R is approximately -
Correct Answer: 100
Explanation:
Ans. (100)
In half wave rectifier $E_{rms} = \int_{0}^{\frac{T}{2}}\frac{E_{0}\sin\omega t dt}{\int_{0}^{T}dt} = \frac{E_{0}}{2} = \frac{200}{2} = 100V$
In half wave rectifier $E_{rms} = \int_{0}^{\frac{T}{2}}\frac{E_{0}\sin\omega t dt}{\int_{0}^{T}dt} = \frac{E_{0}}{2} = \frac{200}{2} = 100V$
Q117
Allen
6. JEE Main Practice Paper
Numerical
For the following circuit and given inputs A and B, find output between $t_{5}$ to $t_{6}$ .
Correct Answer: 1
Explanation:
Ans. (1)
$Y = \overline {{\overline {{A}} \cdot B}} = A + \overline {{B}}$
$Y = \overline {{\overline {{A}} \cdot B}} = A + \overline {{B}}$
Q118
Allen
6. JEE Main Practice Paper
Numerical
The logic operations performed by the given digital circuit.
If $A = 1$ , $B = 1$ , find $Y$ .
Correct Answer: 1
Explanation:
Ans. (1)
Q119
Allen
6. JEE Main Practice Paper
Numerical
In connection with the circuit drawn below, the value of current flowing through 2 kΩ resistor is ____ × 10 $^{-4}$ A.
Correct Answer: 25
Explanation:
Ans. (25)
Current through $2k\Omega$ resistance
$\begin{array}{l} {I = \frac {5}{2 \times 1 0 ^ {3}} = 2. 5 \times 1 0 ^ {- 3} A} \\ {I = 2 5 \times 1 0 ^ {- 4} A} \end{array}$
Current through $2k\Omega$ resistance
$\begin{array}{l} {I = \frac {5}{2 \times 1 0 ^ {3}} = 2. 5 \times 1 0 ^ {- 3} A} \\ {I = 2 5 \times 1 0 ^ {- 4} A} \end{array}$
Q120
Allen
8. JEE Advanced Practice Paper
MCQ
A wire has a mass $0.3 \pm 0.003$ g, radius $0.5 \pm 0.005$ mm and length $6 \pm 0.06$ cm. The maximum percentage error in the measurement of its density is :-
A.
1
B.
2
C.
3
D.
4
Q121
Allen
8. JEE Advanced Practice Paper
MCQ
A vernier callipers having 1 main scale division = 0.1 cm is designed to have a least count of 0.02 cm. If n be the number of divisions on vernier scale and m be the length of vernier scale, then:
A.
n = 10, m = 0.5 cm
B.
n = 9, m = 0.4 cm
C.
n = 10, m = 0.8 cm
D.
n = 10, m = 0.2 cm
Q122
Allen
8. JEE Advanced Practice Paper
MCQ
When the gap is closed without placing any object in the screw gauge whose least count is 0.005 mm, the $5^{th}$ division on its circular scale coincides with the reference line on main scale, and when a small sphere is placed reading on main scale advances by 4 divisions, whereas circular scale reading advances to five times to the corresponding reading when no object was placed. There are 200 divisions on the circular scale. The radius of the sphere is
A.
4.10 mm
B.
4.05 mm
C.
2.10 mm
D.
2.05 mm
Q123
Allen
8. JEE Advanced Practice Paper
MCQ
In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decide to balance a resistance of 4X against Y -
A.
50 cm
B.
80 cm
C.
40 cm
D.
70 cm
Q124
Allen
8. JEE Advanced Practice Paper
MCQ
A student performs an experiment for determination of $g\left(=\frac{4\pi^{2}l}{T^{2}}\right)l\approx1m$ and he commits an error of $\Delta l$ . For the experiment takes the time of n oscillations with the stop watch of least count $\Delta T$ and he commits a human error of 0.1sec. For which of the following data, the measurement of g will be most accurate?
A.
ΔI-5 mm, ΔT-0.2 sec, n-10, Amplitude of oscillation-5mm
B.
ΔI-5 mm, ΔT-0.2 sec, n-20, Amplitude of oscillation-5mm
C.
ΔI-5 mm, ΔT-0.1 sec, n-20, Amplitude of oscillation-1mm
D.
ΔI-1 mm, ΔT-0.1 sec, n-50, Amplitude of oscillation-1mm
Q125
Allen
8. JEE Advanced Practice Paper
MCQ
In an experiment to determine the focal length $(f)$ of a concave mirror by the u-v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then,
A.
$x < f$
B.
$f < x < 2f$
C.
$x = 2f$
D.
$x > 2f$