1

### JEE Main 2016 (Online) 9th April Morning Slot

In the following ‘I’ refers to current and other symbols have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity :
A
ML$-$3 T$-$3 I2
B
M$-$1 L3 T3 I
C
M$-$1 L$-$3 T3 I2
D
M$-$1 L$-$3 T3 I

## Explanation

We know. resistivity ($\rho$) = ${{RA} \over L}$

and conductivity = ${1 \over \rho }$ = ${1 \over {RA}}$

As   R = ${V \over {\rm I}}$

$\therefore$   conductivity = ${{L{\rm I}} \over {VA}}$

Also  V = ${\omega \over q}$ = ${\omega \over {it}}$ = ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {\rm I} \right]\left[ T \right]}}$ = $\left[ {M{L^2}{T^{ - 3}}{{\rm I}^{ - 1}}} \right]$

$\therefore$   Conductivity = ${{\left[ {\rm{L}} \right]\left[ {\rm I} \right]} \over {\left[ {M{L^2}{T^{ - 3}}{{\rm{I}}^{ - 1}}} \right]\left[ {{L^2}} \right]}}$

=   $\left[ {{M^{ - 1}}{L^{ - 3}}{T^3}{{\rm I}^2}} \right]$
2

### JEE Main 2017 (Offline)

The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 $\times$ 10-2 m
rise of water, h = 1.45 $\times$ 10-2m
Using g = 9.80 m/s2 and the simplified relation T = ${{rhg} \over 2} \times {10^3}N/m$, the possible error in surface tension is closest to :
A
10 %
B
0.15 %
C
1.5 %
D
2.4 %

## Explanation

Surface tension,

T = ${{rhg} \over 2} \times {10^3}N/m$

Relative error,

${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$

Percentage error,

${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$

${{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100$

= (0.8 + 0.689) = 1.489 % = 1.5 %
3

### JEE Main 2017 (Online) 8th April Morning Slot

Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be :
A
[M] = [T$-$1 C$-$2 h]
B
[M] = [T$-$1 C2 h]
C
[M] = [T$-$1 C$-$2 h$-$1]
D
[M] = [T C$-$2 h]

## Explanation

Let,

M $\propto$ Tx Cy hz

$\therefore\,\,\,$ [M1LoTo]  =  [T1]x  [L1 T$-$1]y  [M1L2T$-$1]z

[M1Lo To]  =  [Mz Ly + 2z Tx$-$y$-$z]

By comparing both sides we get,

z = 1

y + 2z = 0

x $-$ y $-$ z = 0

$\therefore\,\,\,$ y = $-$ 2z = $-$ 2

x = y + z = $-$2 + 1 = $-$1

$\therefore\,\,\,$ [M] = [M$-$1 C$-$2 h1]
4

### JEE Main 2017 (Online) 9th April Morning Slot

A physical quantity P is described by the relation

P = a$^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}$ b2 c3 d$-$4

If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :
A
8%
B
12%
C
32%
D
25%

## Explanation

Given,

P = a$^{{1 \over 2}}$ b2 c3 d$-$4

Relative error =

${{\Delta P} \over P}$ $\times$ 100 = (${1 \over 2}$ $\times$ ${{\Delta a} \over a}$ + 2${{\Delta b} \over b}$ + 3${{\Delta c} \over c}$ + 4${{\Delta d} \over d}$) $\times$ 100

= ${1 \over 2}$ $\times$ 2 + 2 $\times$ 1 + 3 $\times$ 3 + 4 $\times$ 5

= 32%