1

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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, $$[P] =$$ $$\left[ {{\alpha \over \beta }} \right]$$

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

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

Answer (C)
3

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

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

Answer (A)
4

IIT-JEE 2001 Screening

MCQ (Single Correct Answer)
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.

Answer (D)

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