1.The unit of thermal conductivity is. (NEET 2019)
(a) $ Wm^{{}^-1}K^{{}^-1}$
(b) $Jm\;K^{-1}$
(c) $WmK^{-1}$
(d) $Jm^{{}^-1}K^{{}^-1}$
Ans:- $K=\frac{Qx}{A(T_1-T_2)\ast t}$
where Q is the amount of heat flow, x is the thickness of the slab, A is the area of cross- section, and t is the time taken.
$K=\frac{Jm}{m^2\;Ks}=W\frac1m\frac1K=Wm^{-1}K^{-1}\\$
2.The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are (2012)
(a) $kgms$
(b) $kgms ^ – 2$
(c) $kgs^{-1}$
(d) $kgs$
Ans:-
(c): Damping force, F proportional to v or F = kv where k is the constant of proportionality
$k=\frac Fv=\frac N{ms^{-1}} =(kgms^{-2})/(ms^{-1})=kgs^{-1}$
3.The unit of permittivity of free space, epsilon_{0} is (2004)
(a) coulomb/newton-metre
(b) newton-metre²/coulomb
(c) coulomb²/newton-metre²
(d) coulomb/(newton-metre)²
Ans:-
(c): Force between two charges
$F=\frac1{4\pi\varepsilon_0}\frac{q^2}{r^2}\Rightarrow\varepsilon_0=\frac1{4\pi}\frac{q^2}{Fr^2}=\frac{C^2}{Nm^2}$
4.A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is (NEET 2020)
(a) 0.01 mm
(b) 0.25 mm
(c) 0.5 mm
(d) 1.0 mm
Ans:-
(c) Given: least count = 0.01 and number of circular scale divisions = 50.
Pitch =L. C *No of circular scale division = 0.01 * 50 = 0.5mm .
5.In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where $X=\frac{A^2B^{\displaystyle\frac12}}{C^\frac13D^3}$ will be (NEET 2019)
(a) 10%
(b) (3/13)%
(c) 16%
(d) -10%
Ans:-
$X=\frac{A^2B^{\displaystyle\frac12}}{C^\frac13D^3}$
Maximum percentage error in X
$\left(\frac{dX}X\right)\times100=\left(2\frac{dA}A+\frac12\frac{dB}B+\frac13\frac{dC}C+3\frac{dD}D\right)\times100$
$2\times1+\frac12\times2+\frac13\times3+3\times4=16\%$
6.The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n-1) divisions of main scale. The least count of the vernier callipers is (Odisha NEET 2019)
(a$\frac1{\left(n+1\right)\left(n-1\right)}$ cm
(b$\frac1n$ cm
(c) $\frac1{n^2}$cm
(d) $\frac1{n\left(n+1\right)$cm
Ans:-
(c) If n divisions of vernier scale coincides with (n – 1) divisions of main scale.
Therefore, n VSD = (n – 1) MSD
$\Rightarrow1VSD\;=\frac{\;(n\;-\;1)}n\;MSD$
Least count = 1MSD – VSD = $1MSD\;-\frac{\;\left(n\;-\;1\right)}n\;MSD$
$=\;1MSD\;-\;MSD\;+\frac{\;1}n\times\;MSD\;=\;\frac{\;1}n\times\;MSD$
$=\frac1n\times\frac1n=\frac1{n^2}cm\;\left[1MSD=\frac1ncm\right]$
7.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 levelIf screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is (NEET 2018)
(a) 0.521 cm
(c) 0.053 cm
(b) 0.525 cm
(d) 0.529 cm
Ans:-
(d): Diameter of the ball
= MSR + CSR(Least count) – Zero error
= 5mm + 25 * 0.001 cm – (-0.004) cm = 0.5cm + 25 * 0.001cm – (- 0.004) * cm = 0.529cm .
8.In an experiment, four quantities a, b, c and d are measured with percentage error 1%, 2% 3% and 4% respectively. Quantity P is calculated as follows % error in P is $ P=\frac{a^3b^2}{cd}$ (NEET 2013)
(a) 7%
(b) 4%
(c) 14%
(d) 10%
Ans:-
(c): As % error in P $ P=\frac{a^3b^2}{cd}$
$\frac{\triangle P}P\times100=\left[3\left(\frac{\triangle a}a\right)+2\left(\frac{\triangle b}b\right)+\frac{\triangle c}c+\frac{\triangle d}d\right]\times100$
$=\lbrack3\times1\%+2\times2\%+3\%+4\%\rbrack=14\%$
9.A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e_{1} and dot e_{gamma} respectively, the percentage error in the estimation of g is (Mains 2010)
(a) e_{2} – e
(b) e_{1} + 2e_{2}
(c) e_{1} + e_{2}
(d) e_{1} – 2e_{2}
Ans:-
(b): From the relation, $ h=ut+\frac12gt^2$
$ h=\frac12gt^2\Rightarrow g=\frac{2h}{t^2} $( . body initially at rest)
Taking natural logarithm on both sides, we get
$ \ln(g)\;=\;\ln(h)\;-\;2\;\ln(t)\;$
Differentiating,
$ \frac{\triangle g}g=\frac{\triangle h}h=-2\frac{\triangle t}t$
For maximum permissible error,
or $ \left(\frac{\triangle g}g\times100\right)=\left(\frac{\triangle h}h\times100\right)=-2\left(\frac{\triangle t}t\times100\right)$
According to problem
$ \frac{\triangle h}h\times100=e_1\;and\;\frac{\triangle t}t\times100=e_2$
Therefore, ${\left(\frac{\triangle g}g\times100\right)}_{max}=e_1+2e_2$
- If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be (2008)
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Ans:-
(d): $ V=\frac43\pi R^3\;;\;InV=ln(\frac43\pi)+ln(R^3)$
Differentiating, $ \frac{dV}V=3\frac{dR}R$
Error in the determination of the volume
$=3\times2\%=6\%$
