1

IIT-JEE 2006

A student performs an experiment for determination of $\mathrm g\left(=\frac{4\mathrm\pi^2\mathcal l}{\mathrm T^2}\right)$. The error in length $\mathcal l$ is $\triangle\mathcal l$ and in the time T is $\triangle\mathrm T$ and n is number of times the reading is taken.The reading of g is most accurate for
A
$\begin{array}{l}\triangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\sec\;\;\;\;\;\;\;\;\;\;\;\;10\end{array}$
B
$\begin{array}{l}\triangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\sec\;\;\;\;\;\;\;\;\;\;\;\;20\end{array}$
C
$\begin{array}{l}\triangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\sec\;\;\;\;\;\;\;\;\;\;\;\;10\end{array}$
D
$\begin{array}{l}\triangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\triangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\\1\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\sec\;\;\;\;\;\;\;\;\;\;\;\;50\end{array}$
2

IIT-JEE 2005 Screening

Which of the following set have different dimensions?
A
Pressure, Young's modulus, Stress
B
EMF, Potential difference, Electric potential
C
Heat, Work done, Energy
D
Dipole moment, Electric flux, Electric field
3

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
4

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

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

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