1

### IIT-JEE 2004 Screening

A wire of length $l = 6 \pm 0.06$ cm and $r = 0.5 \pm 0.005$ cm and mass $m = 0.3 \pm 0.003$ gm. Maximum percentage error in density is
A
4
B
2
C
1
D
6.8
2

### IIT-JEE 2004 Screening

Pressure depends on distance as, $P = {\alpha \over \beta }\exp \left( { - {{\alpha z} \over {k\theta }}} \right)$, where $\alpha$, $\beta$ are constants, z in distance, k is Boltzman's constant and $\theta$ is temperature. The dimention of $\beta$ are
A
[M0L0T0]
B
[M-1L-1T-1]
C
[M0L2T0]
D
[M-1L1T2]

## Explanation

The given expression is

$P = {\alpha \over \beta }\exp \left( { - {{\alpha z} \over {k\theta }}} \right)$

Now,

$[z] = [{L^1}]$

$[\theta ] = [{K^1}]$

$[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

Since the terms which are raised to power in an exponential function are dimensionless, we get

$[\alpha] = \left[ {{{k\theta } \over z}} \right] = [{M^1}{L^1}{T^{ - 2}}]$

Also, $[P] =$ $\left[ {{\alpha \over \beta }} \right]$

$\Rightarrow$ $\left[ \beta \right] = \left[ {{\alpha \over P}} \right]$

$\Rightarrow$ $\left[ \beta \right] = \left[ {{{{M^1}{L^1}{T^{ - 2}}} \over {M{L^{ - 1}}{T^{ - 2}}}}} \right]$ = $[{M^0}{L^2}{T^0}]$

3

### IIT-JEE 2003 Screening

A cube has a side of length 1.2 ✕ 10-2 m. Calculate its volume.

A
1.7 ✕ 10-6 m3
B
1.73 ✕ 10-6 m3
C
1.70 ✕ 10-6 m3
D
1.732 ✕ 10-6 m3

## Explanation

We have

V = l3 = (1.2 $\times$ 10$-$2)3 = 1.728 $\times$ 10$-$6 m3

Since l has two significant figures, V should also have two significant figures; hence,

V = 1.7 $\times$ 10$-$6 m3

4

### IIT-JEE 2001 Screening

A quantity X is given by ${\varepsilon _0}L{{\Delta V} \over {\Delta t}}$ where ${\varepsilon _0}$ is the permittivity of the free space, L is a length, ${\Delta V}$ is a potential difference and ${\Delta t}$ is a time interval. The dimentional formula for X is the same as that of
A
resistance
B
charge
C
voltage
D
current

## Explanation

Given that $X = {\varepsilon _0}L\left( {{{\Delta V} \over {\Delta t}}} \right)$

Therefore, dimensional formula for X is

$[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}][L]{{[M{L^2}{T^{ - 3}}{I^{ - 1}}]} \over {[T]}} = [I]$

Thus, dimensional formula for X is same as that of current.