Units & Measurements — Page 7 | JEE Physics

Q61

The quantities x =

{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

, y =

{E \over B}

and z =

{l \over {CR}}

are defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and

{{\varepsilon _0}}

,

{{\mu _0}}

, - free space permittivity and permeability respectively. Then :

A Only y and z have the same dimension

B x, y and z have the same dimension

C Only x and y have the same dimension

D Only x and z have the same dimension

Correct Answer

Option B

Solution

x =

{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

= speed $\therefore$ [x] = [L1T–1] y =

{E \over B}

= speed $\therefore$ [y] = [L1T–1] z =

{l \over {CR}}

=

{l \over \tau }

$\Rightarrow$ [z] = [L1T–1] So, x, y, z all have the same dimensions.

Q62

A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively :

A Positive, 0.1 mm

B Positive, 0.1

\mu

m

C Positive, 10

\mu

m

D Negative, 2

\mu

m

Correct Answer

Option C

Solution

Least count of screw gauge =

{{0.5} \over {50}}

= 1 $\times$ 10-5 m = 10 $\mu$ m Zero error in positive.

Q63

A quantity f is given by

f = \sqrt {{{h{c^5}} \over G}}

where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of :

A Energy

B Momentum

C Area

D Volume

Correct Answer

Option A

Solution

[h] = M1L2T–1 [C] = L1T–1 [G] = M–1L3T–2 [f] =

\sqrt {{{M{L^2}{T^{ - 1}} \times {L^5}{T^{ - 5}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}}}

= M1L2T–2

Q64

If mass is written as

m=k \mathrm{c}^{\mathrm{P}} G^{-1 / 2} h^{1 / 2}

then the value of

P

will be : (Constants have their usual meaning with

k a

dimensionless constant)

A 1/3

B

-

1/3

C 1/2

D 2

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\ & \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2} \end{aligned}

By comparing

\mathrm{P}=1 / 2

Q65

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as :

A (5.54

\pm

0.07) mm

B (5.5375

\pm

0.0740) mm

C (5.5375

\pm

0.0739) mm

D (5.538

\pm

0.074) mm

Correct Answer

Option A

Solution

Given, dav = 5.5375 mm

\Delta

d = 0.07395 mm Significant rule says that reading should has same significant figure as that of reading given.

$\because$ Measured data are up to two digits after decimal.

$\therefore$ 5.5375 rounded to $\to$ 5.54

Q66

If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is

A [P2AT–2]

B

\left[ {{P^{{1 \over 2}}}A{T^{ - 1}}} \right]

C

\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]

D [PA–1T–2]

Correct Answer

Option C

Solution

Let [E] = K[P]x[A]y [T]z [ML2T–2] = [MLT–1]x[L2]y[T]z [ML2T–2] = [Mx][Lx+2y][T–x+z] Comparing both side we get, x = 1 x + 2y = 2 $\Rightarrow$ 1 + 2y= 2 or y =

\frac{1}{2}

z – x = –2 $\Rightarrow$ z–1 = –2 or z = –1 $\therefore$ [E] =

\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]

Q67

Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

A 2.123 cm

B 2.124 cm

C 2.125 cm

D 2.121 cm

Correct Answer

Option B

Solution

Using a screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured as: Measurement = (Main scale reading) + (Circular scale reading × Least count) where the least count is calculated as the pitch of the screw gauge divided by the number of divisions on the circular scale: Least count = (Pitch of screw gauge) / (Number of circular scale divisions) Least count =

{{0.1} \over {50}}

= 0.002 cm Now if we multiply division of circular scale with least count then we get 0th digit of fraction part even.

Here only option B has 0th digit of fraction part even.

Q68

The dimension of

{{{B^2}} \over {2{\mu _0}}}

, where B is magnetic field and

{{\mu _0}}

is the magnetic permeability of vacuum, is :

A ML2T–2

B MLT–2

C ML-1T–2

D ML2T–1

Correct Answer

Option C

Solution

As

{{{B^2}} \over {2{\mu _0}}}

= Energy per unit volume $\therefore$ Dimension =

{{M{L^2}{T^{ - 2}}} \over {{L^3}}}

= ML-1T–2

Q69

The dimension of stopping potential V0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :

A h1/3 G2/3 c1/3 A–1

B h0 c5 G-1 A-1

C h2/3 c5/3 G1/3 A–1

D h2 G3/2 c1/3 A–1

Correct Answer

Option B

Solution

V0 $\propto$ hPcQGRIS [V0] = [M1L2T–3A–1] [c] = [L1T–1] [h] = [M1L2T–1] [G] = [M–1L3T–2] [I] = [A] $\therefore$ [M1L2T–3A–1] = [MP–R L2P+Q+3R T–P–Q–2R AS] Comparing dimension of M, L, T, A, we get P – R = 1 ; 2P + Q + 3R = 2 – P – Q – 2R = – 3 ; S = – 1 $\Rightarrow$ P = 0, Q = 5, R = –1, S = –1 $\therefore$ V0 $\propto$ h0 c5 G-1 A-1

Q70

If the screw on a screw-gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is :

A 0.001 mm

B 0.01 cm

C 0.02 mm

D 0.001 cm

Correct Answer

Option D

Solution

Pitch =

{3 \over 6}

mm = 0.5 mm Least count =

{{0.5} \over {50}}

mm =

{1 \over {100}}

= 0.01 mm = 0.001 cm