Units And Dimensions(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

UNITS AND DIMENSIONS MCQs

Total Questions : 40 | Page 1 of 4 pages

Question 1.

If radius of the sphere is (5.3 $\pm$ 0.1)cm. Then percentage error in its volume will be

3+6.01 $\times \frac{100}{5.3}$
$\frac{1}{3} \times$ 0.01 $\times \frac{100}{5.3}$
$(\frac{3 \times 0.1}{5.3}) \times 100$
$\frac{0.1}{5.3} \times 100$

Discuss Question

Answer: Option C. ->

(\frac{3 \times 0.1}{5.3}) \times 100

:
C

Volume of sphere (v) = $\frac{4}{3} π γ^{3}$
Percentage error in volume =3 $\times \frac{Δ γ}{γ} \times$ 100 = (3 $\times \frac{0.1}{5.3}) \times$ 100

Question 2.

The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is

Discuss Question

Answer: Option D. -> 8%
:
D

$P = \frac{F}{A} = \frac{F}{l^{2}},$ So maximum error in pressure (P)
$(\frac{Δ P}{P} \times 100)_{m a x} = \frac{Δ F}{F} \times 100 + 2 \frac{Δ l}{1} \times 100$

= 4 % + 2 \times 2 % = 8 %

Question 3.

The position of a particle at time t is given by the relation x(t)= $(\frac{v_{0}}{α}) (1 - c^{- a t})$ where $v_{0}$ and $α$ are constants . The dimensions of $v_{0}$ and $α$ are respectively

$M^{0} L T^{- 1}$ and $T^{- 1}$
$M^{0} L^{1} T^{0}$ and $T^{- 1}$
$M^{0} L^{1} T^{- 1}$ and $L T^{- 2}$
$M^{0} L^{1} T^{- 1}$ and T

Discuss Question

Answer: Option A. ->

M^{0} L T^{- 1}

and

T^{- 1}

:
A

Dimension of $α$ t = $[M^{0} L^{0} T^{0}] ∴ [α] = [T^{- 1}]$
Again $[\frac{v_{0}}{α}]$ = [L] so $[v_{0}] = [L T^{- 1}]$

Question 4.

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation $g = 4 π^{2} (\frac{l}{T^{2}})$ will be

Discuss Question

Answer: Option C. -> 7%
:
C

Percentage error in $g = (% e r r o r i n l) + 2 (% e r r o r i n T) = 1 % + 2 (3 %) = 7 %$

Question 5.

The length, breadth, and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm The volume of the block according to the idea of significant figures should be

1 $\times 10^{2} c m^{3}$
2 $\times 10^{2} c m^{3}$
1.763 $\times 10^{2} c m^{3}$
None of these

Discuss Question

Answer: Option B. -> 2

\times 10^{2} c m^{3}

:
B

Volume V = l $\times$ b $\times$ t

= 12 $\times$ 6 $\times$ 2.45 = 176.4 $c m^{3}$

V = 176.4 $\times 10^{2} c m^{3}$

Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V = 2 $\times 10^{2} c m^{3} .$

Question 6.

The velocity v (in cm / sec) of a particle is given in terms of time t(in sec) by the relation v = $α$ t + $\frac{b}{t + c}$ ; the dimensions of a, b and c are

a = $L^{2}$ , b = T, c = $L T^{2}$
a = $L T^{2}$ , b = LT, c = L
a = $L T^{- 2}$ , b = L, c = T
a = L, b = LT, c = $T^{2}$

Discuss Question

Answer: Option C. -> a =

L T^{- 2}

, b = L, c = T
:
C

From the principle of dimensional homogeneity [v] = $[α t] \Rightarrow [α] = [L T^{- 2}]$ . similarly [b] = [L] and [c] = [T]

Question 7.

From the equation $t a n θ = \frac{r g}{v^{2}}$ , one can obtain the angle of banking $θ$ for a cyclist taking a curve (the symbols have their usual meanings). Then say, it is

Both dimensionally and numerically correct
Neither numerically nor dimensionally correct
Dimensionally correct only
Numerically correct only

Discuss Question

Answer: Option C. -> Dimensionally correct only
:
C

Given equation is dimensionally correct because both sides are dimensionless but numerically wrong because the correct equation is $t a n θ = \frac{v^{2}}{r g} .$

Question 8.

The SI unit of universal gas constant (R) is

Watt $K^{- 1} m o l^{- 1}$
Newton $K^{- 1} m o l^{- 1}$
Joule $K^{- 1} m o l^{- 1}$
Erg $K^{- 1} m o l^{- 1}$

Discuss Question

Answer: Option C. -> Joule

K^{- 1} m o l^{- 1}

:
C

PV = nRT $\Rightarrow$ R = $\frac{P V}{n T}$ = $\frac{J o u l e}{m o l e \times K e l v i n} = J K^{- 1} m o l^{- 1}$

Question 9.

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity $η$ flowing per second through a tube of radius r and length l and having a pressure difference P across its end, is