1

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]
2

JEE Main 2017 (Online) 9th April Morning Slot

A physical quantity P is described by the relation

P = a$^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}$ 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%
3

JEE Main 2018 (Offline)

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
A
6%
B
2.5%
C
3.5%
D
4.5%

Explanation

Density of a material (d) = ${M \over {{L^3}}}$

$\therefore$ Error in density,${{\Delta d} \over d} = {{\Delta M} \over M} + 3{{\Delta L} \over L}$

${{\Delta d} \over d} \times 100 = {{\Delta M} \over M} \times 100 + 3{{\Delta L} \over L} \times 100$

$\Rightarrow {{\Delta d} \over d} \times 100 = 1.5\% + 3\left( 1 \right)\%$ = 4.5 %
4

JEE Main 2018 (Offline)

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 $\Omega$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.
A
2.5 $\Omega$
B
1 $\Omega$
C
1.5 $\Omega$
D
2 $\Omega$

Explanation

Internal resistance of potentiometer,

r = $\left( {{E \over V} - 1} \right) \times R$

Initially when no current passes through the galvanometer then

emf, E = K (52)

After cell is shunted by a resistance 5 $\Omega$, then,

Terminal voltage, V = K(40)

$\therefore\,\,\,$ r = $\left( {{{52K} \over {40K}} - 1} \right)$ $\times$ 5

= $\left( {{{26} \over {20}} - 1} \right)$ $\times$ 5

= 1.5 $\Omega$