1. The estimate of absolute error in the measurement of time using a clock is
\(10^{-2}~\text{s}.\) The time difference,
\(t=t_1-t_2,\) between two events is determined by using the clock. The error in
\(t\) is:
1. |
\(10^{-2}~\text{s}\) |
2. |
\(2\times10^{-2}~\text{s}\) |
3. |
\({\Large\frac12}\times10^{-2}~\text{s}\) |
4. |
zero |
2. The percentage errors in the measurement of mass and momentum of an object are \(1\%\) and \(2\%\) respectively. The percentage error in the measurement of kinetic energy of the object will be:
1. \(1\%\)
2. \(3\%\)
3. \(4\%\)
4. \(5\%\)
3. The dimension of which group of quantities is the same?
\(h\) : Planck's constant, \(K\) : kinetic energy, \(\omega\) : angular speed/frequency, \(F\): force, \(L\) : inductance, \(i\) : current, \(q\) : charge, \(t\) : time, \(x\): distance
1. \(h, ~Ftx,~ Liq\)
2. \(K , h\omega , \omega Li\)
3. \(Fx, ~Li^2,~K \omega\)
4. \(\dfrac{Fx}{t} ,~ Kx, ~ ht\)
4. Among the following numerical values, which one has three significant figures?
1. \(0.300\)
2. \(30.30\)
3. \(0.030\)
4. \(3.033\)
5. A student measured the diameter of a small steel ball using a screw gauge of least count
\(0.001\) cm. The main scale reading is
\(5\) mm and zero of circular scale division coincides with
\(25\) divisions above the reference level. If the screw gauge has a zero error of
\(-0.004\) cm, the correct diameter of the ball is:
1. |
\(0.521\) cm |
2. |
\(0.525\) cm |
3. |
\(0.053\) cm |
4. |
\(0.529\) cm |
6. The quantities which have the same dimensions as those of solid angle are:
1. stress and angle
2. strain and arc
3. angular speed and stress
4. strain and angle
7. A vernier calipers has
\(1~\text{mm}\) marks on the main scale. It has
\(20\) equal divisions on the vernier scale which match with
\(16\) main scale divisions. The least count of this vernier calipers is:
1. |
\(0.02~\text{mm}\) |
2. |
\(0.05~\text{mm}\) |
3. |
\(0.10~\text{mm}\) |
4. |
\(0.20~\text{mm}\) |
8. A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: \(0\) mm
Circular scale reading: \(52\) divisions
Given that \(1\) mm on the main scale corresponds to \(100\) divisions on the circular scale, the diameter of the wire that can be inferred from the given data is:
1. |
\(0.26\) cm |
2. |
\(0.052\) cm |
3. |
\(0.52\) cm |
4. |
\(0.026\) cm |
9. Match the units mentioned in
List-I with the units in
List-II.
|
List-I |
|
List-II |
(a) |
H/s |
(i) |
s2 |
(b) |
H×A |
(ii) |
Wb |
(c) |
H×F |
(iii) |
s |
(d) |
\(\Omega\)×F |
(iv) |
\(\Omega\) |
1. |
(a)–(ii), (b)–(iv), (c)–(i), (d)–(iii) |
2. |
(a)–(iv), (b)–(ii), (c)–(i), (d)–(iii) |
3. |
(a)–(iv), (b)–(ii), (c)–(iii), (d)–(i) |
4. |
(a)–(ii), (b)–(iii), (c)–(iv), (d)–(i) |
10. Using a screw gauge with pitch \(0.1 ~\text{cm}\) and \(50\) divisions on its circular scale, the thickness of an object is measured. It should be accurately recorded as:
1. \(2.124~\text{cm}\)
2. \(2.121~\text{cm}\)
3. \(2.125~\text{cm}\)
4. \(2.123~\text{cm}\)
11. The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are:
1. |
Random errors |
2. |
Instrumental errors |
3. |
Personal errors |
4. |
Least count errors |
12. The least count of a stopwatch is \(0.1\) sec. The time of \(20\) oscillations of the pendulum is found to be \(20\) sec. The percentage error in the time period is:
1. \(0.25\%\)
2. \(0.5\%\)
3. \(0.75\%\)
4. \(1.0\%\)
13. Match
List-I with
List-II.
List-I |
List-II |
(a) |
Gravitational constant (\(G\)) |
(i) |
\([{L}^2 {~T}^{-2}] \) |
(b) |
Gravitational potential energy |
(ii) |
\([{M}^{-1} {~L}^3 {~T}^{-2}] \) |
(c) |
Gravitational potential |
(iii) |
\([{LT}^{-2}] \) |
(d) |
Gravitational intensity |
(iv) |
\([{ML}^2 {~T}^{-2}]\) |
Choose the correct answer from the options given below:
|
(a) |
(b) |
(c) |
(d) |
1. |
(iv) |
(ii) |
(i) |
(iii) |
2. |
(ii) |
(i) |
(iv) |
(iii) |
3. |
(ii) |
(iv) |
(i) |
(iii) |
4. |
(ii) |
(iv) |
(iii) |
(i) |
14. Taking into account the significant figures, what is the value of \((9.99~\text{m}-0.0099~\text{m})\)?
1. \(9.98~\text{m}\)
2. \(9.980~\text{m}\)
3. \(9.9~\text{m}\)
4. \(9.9801~\text{m}\)
15. Express the result of the calculation to \(3\) significant figures:
\(2\times0.536+0.0050+2.100\)
1. \(3.177\)
2. \(3.18\)
3. \(3.19\)
4. \(3.2\)
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