Units & Measurements — Page 5 | NEET Physics

Q41

A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is :

A 0.052 cm

B 0.52 cm

C 0.026 cm

D 0.26 cm

Correct Answer

Option A

Solution

L.C. =

{{Pitch} \\over {CSD}}

= {{1mm} \\over {100}}

= 0.01 m = 0.001 cm Radius = M.S. + CSR(L.C) = 0 + 52 (0.001) = 0.052 cm

Q42

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

A [F] [A

-

1 ] [T]

B [F] [A] [T]

C [F] [A] [T 2 ]

D [F] [A] [T

-

1 ]

Correct Answer

Option C

Solution

E $\\propto$ F a A b T c [M 1 L 2 T $-$ 2 ] $\\propto$ [M 1 L 1 T $-$ 2 ] a [LT $-$ 2 ] b [T] c [ML 2 T –2 ] $\\propto$ [M a L a + b T –2a – 2b + c ] a = 1 a + b = 2 $\\Rightarrow$ b = 1 $-$ 2a $-$ 2b + c = $-$ 2 $\\Rightarrow$ c = 2 a = 1, b = 1, c = 2 E $\\propto$ [F] [A] [T 2 ]

Q43

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is :

A 0.25 mm

B 0.5 mm

C 1.0 mm

D 0.01 mm

Correct Answer

Option B

Solution

Least count of screw gauge =

\\frac{Pitch}{No\\ of\\ divisions\\ on\\ circular\\ scale}

0.01 mm =

{{Pitch} \\over {50}}

Pitch = 0.5 mm

Q44

Taking into account of the significant figures, what is the value of

9.99 m - 0.0099m

?

A 9.98 m

B 9.980 m

C 9.9 m

D 9.9801 m

Correct Answer

Option A

Solution

9.99 - 0.0099

= 9.9801 m Here, 9.98 is having the least number of decimal places of two, so answer should also have only two decimal places.

So answer is 9.98 m.

Q45

Dimensions of stress are :

A

\left[ {M{L^{ 2}}{T^{ - 2}}} \right]

B

\left[ {M{L^{ 0}}{T^{ - 2}}} \right]

C

\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]

D

\left[ {M{L^{}}{T^{ - 2}}} \right]

Correct Answer

Option C

Solution

Stress = {{Force} \\over {Area}}

= \\left[ {{{ML{T^{ - 2}}} \\over {{L^2}}}} \\right] = \\left[ {M{L^{ - 1}}{T^{ - 2}}} \\right]

Q46

The energy required to break one bond in DNA is 10 -20 J. This value in eV is nearly :

A 0.6

B 0.06

C 0.006

D 6

Correct Answer

Option B

Solution

1eV = 1.6 \\times {10^{ - 19}}J

1J = {1 \\over {1.6 \\times {{10}^{ - 19}}}}eV

{10^{ - 20}}J = {{{{10}^{ - 20}}} \\over {1.6 \\times {{10}^{ - 19}}}}eV

= 0.06 eV

Q47

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2 %, 3 %and 4 % respectively. Then the maximum percentage of error in the measurement X, where X =

{{{A^2}{B^{1/2}}} \\over {{C^{1/3}}{D^3}}}

, will be :

A 16%

B -10%

C 10%

D

\left( {{3 \over {13}}} \right)

%

Correct Answer

Option A

Solution

Given, X =

{{{A^2}{B^{1/2}}} \\over {{C^{1/3}}{D^3}}}

{{\\Delta X} \\over X} = {{2\\Delta A} \\over A} + {1 \\over 2}{{\\Delta B} \\over B} + {1 \\over 3}{{\\Delta C} \\over C} + 3{{\\Delta D} \\over D}

$\\Rightarrow$

{{\\Delta X} \\over X} \\times 100 = 2\\left( {1\\% } \\right) + {1 \\over 2}\\left( {2\\% } \\right) + {1 \\over 3}\\left( {3\\% } \\right) + 3\\left( {4\\% } \\right)

= 16 %

Q48

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

A 0.521 cm

B 0.525 cm

C 0.053 cm

D 0.529 cm

Correct Answer

Option D

Solution

Diameter of the ball = MSR + CSR × (Least count) – Zero error = 5 mm + 25 × 0.001 cm – (–0.004) cm = 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm.

Q49

A physical quantity of the dimensions of length that can be formed out of c, G and

{{{e^2}} \\over {4\\pi {\\varepsilon _0}}}

is [c is velocity of light, G is the universal constant of gravitation and e is charge]

A

{c^2}{\left[ {G - {{{e^2}} \over {4\pi {\varepsilon _0}}}} \right]^{1/2}}

B

{1 \over {{c^2}}}{\left[ {{{{e^2}} \over {G\,4\pi {\varepsilon _0}}}} \right]^{1/2}}

C

{1 \over c}G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}

D

{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}}

Correct Answer

Option D

Solution

Dimension of

{{{e^2}} \\over {4\\pi {\\varepsilon _0}}}

= [ F $\\times$ d 2 ] = [ML 3 T -2 ] Dimension of G = [M -1 L 3 T -2 ], Dimension of c = [LT -1 ] Now assume dimension of length is related as, L = [c] x [G] y [

{{{e^2}} \\over {4\\pi {\\varepsilon _0}}}

] z $\\therefore$ [L 1 ] = [ML 3 T -2 ] z [M -1 L 3 T -2 ] y [LT -1 ] x Comparing both sides and solving we get, x = -2, y =

{1 \\over 2}

, z =

{1 \\over 2}

$\\therefore$ L =

{1 \\over {{c^2}}}{\\left[ {G{{{e^2}} \\over {\\,4\\pi {\\varepsilon _0}}}} \\right]^{1/2}}

Q50

Planck's constant (h), speed of light in vacuum (c) and Newton's gravitional constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length ?

A

{{\sqrt {hG} } \over {{c^{3/2}}}}

B

{{\sqrt {hG} } \over {{c^{5/2}}}}

C

\sqrt {{{hc} \over G}}

D

\sqrt {{{Gc} \over {{h^{3/2}}}}}

Correct Answer

Option A

Solution

According to question, L $\\propto$ h p c q G r [M 0 LT 0 ] = [ML 2 T -1 ] p [LT -1 ] q [M -1 L 3 T -2 ] r Equating power both sides, we get p - r = 0 ......(1) 2p + q + 3r = 1 ......(2) - p - q - 2r = 0 ......(

3) Solving equation (1), (2), (3), we get p = r =

{1 \\over 2}

, q =

- {3 \\over 2}

$\\therefore$ L =

{{\\sqrt {hG} } \\over {{c^{3/2}}}}