JEE Main Units and Measurements PYQs (Set-4) offers a new set of carefully selected previous year questions with detailed solutions to strengthen your understanding of fundamental concepts. This set focuses on important topics such as dimensional analysis, significant figures, and error calculations, helping you improve accuracy and speed. Regular practice of these PYQs will enhance your problem-solving skills and boost your confidence for the JEE Main exam.
JEE Main PYQs Previous Year Questions With Solutions of Chapter Units and Measurements Class 11 Physics
Build a strong foundation by first exploring Class 11 Physics Notes for clear concepts and complete theory.
Next, understand the basics in depth through the Units and Measurements chapter to strengthen your fundamentals.
Once your concepts are clear, practice Units and Measurements JEE Main PYQs Set-1 to get familiar with question patterns.
Then move to Units and Measurements JEE Main PYQs Set-2 to improve accuracy and problem-solving speed.
Further enhance your preparation with Units and Measurements JEE Main PYQs Set-3 for advanced practice and better exam readiness.
Problem : The following observations were taken for determining surface tension $T$ of water by capillary rise method:
Diameter of capillary, $d = 1.25 \times 10^{-2}$ m;
Rise of water, $h = 1.45 \times 10^{-2}$ m.
Using $g = 9.80$ m/s$^2$ and the simplified relation $T = \left( \dfrac{rhg}{2} \right) \times 10^3$ N/m, the possible error in surface tension is closest to:
(a) $1.5\%$
(b) $2.4\%$
(c) $10\%$
(d) $5\%$
[JEE Main 2017]
Solution: Ans. (a)
$T = \dfrac{rhg}{2} \times 10^3 = \dfrac{dhg}{4} \times 10^3$
Relative error in $T$: $\dfrac{\Delta T}{T} = \dfrac{\Delta d}{d} + \dfrac{\Delta h}{h}$
Assuming the least count for $d$ and $h$ is $0.01 \times 10^{-2}$ m:
$\dfrac{\Delta T}{T} = \dfrac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}} + \dfrac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}}$
$\dfrac{\Delta T}{T} = \dfrac{1}{125} + \dfrac{1}{145} = 0.008 + 0.0069 = 0.0149 \approx 1.5\%$.
Problem : In the measurement of the physical quantity $X = \dfrac{A^2 B^{1/2}}{C^{1/3} D^3}$, the percentage error in the measurements of $A, B, C$ and $D$ are $4\%, 2\%, 3\%$ and $1\%$, respectively. Then, the percentage error in $X$ is:
(a) $11\%$
(b) $12\%$
(c) $13\%$
(d) $14\%$
[JEE Main 2016]
Solution: Ans. (c)
The percentage error in $X$ is given by:
$\dfrac{\Delta X}{X} \times 100 = \left[ 2 \left( \dfrac{\Delta A}{A} \right) + \dfrac{1}{2} \left( \dfrac{\Delta B}{B} \right) + \dfrac{1}{3} \left( \dfrac{\Delta C}{C} \right) + 3 \left( \dfrac{\Delta D}{D} \right) \right] \times 100$
$$\dfrac{\Delta X}{X} \times 100 = 2(4\%) + \dfrac{1}{2}(2\%) + \dfrac{1}{3}(3\%) + 3(1\%)$$
$$\dfrac{\Delta X}{X} \times 100 = 8\% + 1\% + 1\% + 3\% = 13\%$.
Problem : A body of mass $m = 3.513 \text{ kg}$ is moving along the x-axis with a speed of $5.00 \text{ ms}^{-1}$. The magnitude of its momentum is recorded as:
(a) $17.6 \text{ kg ms}^{-1}$
(b) $17.565 \text{ kg ms}^{-1}$
(c) $17.56 \text{ kg ms}^{-1}$
(d) $17.57 \text{ kg ms}^{-1}$
[AIEEE 2008]
Solution: Ans. (a)
Momentum $p = mv$
$p = 3.513 \times 5.00 = 17.565 \text{ kg ms}^{-1}$
As the number of significant figures in $5.00$ is 3, the result should also be in 3 significant figures (minimum).
$p = 17.6 \text{ kg ms}^{-1}$
Problem : A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 92s and 95s. If the minimum division in the measuring clock is 1s, then the reported mean time should be:
(a) (92 ± 2)s
(b) (92 ± 5)s
(c) (92 ± 1.8)s
(d) (92 ± 3)s
[JEE Main 2016]
Solution: Ans. (a)
Arithmetic mean time $\bar{x} = \dfrac{\sum x_i}{N}$
$\bar{x} = \dfrac{90 + 91 + 92 + 95}{4} = \dfrac{368}{4} = 92\text{ s}$
Mean absolute deviation $\Delta \bar{x} = \dfrac{\sum |x_i – \bar{x}|}{N}$
$\Delta \bar{x} = \dfrac{|90-92| + |91-92| + |92-92| + |95-92|}{4}$
$\Delta \bar{x} = \dfrac{2 + 1 + 0 + 3}{4} = \dfrac{6}{4} = 1.5\text{ s}$
Since the resolution (minimum division) of the measuring clock is $1\text{ s}$, the error cannot be less than the least count. The mean absolute error ($1.5\text{ s}$) is rounded off to the precision of the instrument, which is $2\text{ s}$ (since it must be reported in terms of the least count or significant figures consistent with the data).
Therefore, the reported mean time is $(92 \pm 2)\text{ s}$.
Problem : A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading : 58.5 degree
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data is:
(a) 58.59 degree
(b) 58.77 degree
(c) 58.65 degree
(d) 59 degree
[AIEEE 2012]
Solution: Ans. (c)
Given:
$1 \text{ Main Scale Division (MSD)} = 0.5^\circ$
Number of Vernier Scale Divisions ($VSD$) $= 30$
$30 \text{ VSD}$ coincides with $29 \text{ MSD}$
$\Rightarrow 1 \text{ VSD} = \frac{29}{30} \text{ MSD}$
Step 1: Calculate the Least Count (LC)
$LC = 1 \text{ MSD} – 1 \text{ VSD}$
$LC = 1 \text{ MSD} – \frac{29}{30} \text{ MSD}$
$LC = \frac{1}{30} \text{ MSD}$
$LC = \frac{1}{30} \times 0.5^\circ = \frac{1}{30} \times \frac{1}{2}^\circ = \frac{1}{60}^\circ$
Step 2: Calculate the Total Reading
Total Reading $= \text{Main Scale Reading (MSR)} + (\text{Vernier Scale Reading} \times LC)$
Reading $= 58.5^\circ + \left( 9 \times \frac{1}{60} \right)^\circ$
Reading $= 58.5^\circ + \left( \frac{3}{20} \right)^\circ$
Reading $= 58.5^\circ + 0.15^\circ$
Reading $= 58.65^\circ$
TOPIC 3 : Dimensions
Problem : Which of the following equations is dimensionally incorrect?
Where, $t = \text{time}$, $h = \text{height}$, $s = \text{surface tension}$, $\theta = \text{angle}$, $\rho = \text{density}$, $a, r = \text{radius}$, $g = \text{acceleration due to gravity}$, $V = \text{volume}$, $p = \text{pressure}$, $W = \text{work done}$, $\tau = \text{torque}$, $\varepsilon = \text{permittivity}$, $E = \text{electric field}$, $J = \text{current density}$, $L = \text{length}$.
(a) $V = \dfrac{\pi p a^4}{8 \eta L}$
(b) $h = \dfrac{2s \cos\theta}{\rho r g}$
(c) $J = \varepsilon \dfrac{\partial E}{\partial t}$
(d) $W = \tau\theta$
[JEE Main 2021, 31 Aug Shift-I]
Solution: Ans. (a)
We know the dimensional formula of Volume ($V$) is $[M^0 L^3 T^0]$.
For option (a), let’s check the dimensions of the RHS: $\dfrac{\pi p a^4}{8 \eta L}$.
From Stokes’ law, $F = 6\pi\eta r v$, so $[\eta] = \dfrac{[F]}{[r][v]} = \dfrac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$.
Dimensions of pressure $p = \dfrac{[F]}{[A]} = [ML^{-1}T^{-2}]$.
Dimensions of $a^4 = [L^4]$ and $L = [L]$.
RHS Dimensions $= \dfrac{[ML^{-1}T^{-2}][L^4]}{[ML^{-1}T^{-1}][L]} = \dfrac{[ML^3 T^{-2}]}{[M T^{-1}]} = [L^3 T^{-1}]$.
Since $[V] = [L^3] \neq [L^3 T^{-1}]$, option (a) is dimensionally incorrect. (Note: This is Poiseuille’s equation for volume flow rate $\dfrac{dV}{dt}$, not Volume $V$).
For option (b): $h = \dfrac{2s \cos\theta}{\rho r g}$
LHS Dimensions $= [L]$.
RHS Dimensions $= \dfrac{[MT^{-2}]}{[ML^{-3}][L][LT^{-2}]} = \dfrac{[MT^{-2}]}{[MT^{-4}L^2]}$ (simplified) $= [L]$.
Hence, (b) is dimensionally correct.
For option (c): $J = \varepsilon \dfrac{\partial E}{\partial t}$
Dimensions of Current Density $J = \dfrac{[I]}{[A]} = [AL^{-2}]$.
We know $D = \varepsilon E$ and $J = \dfrac{\partial D}{\partial t}$ (displacement current density).
Dimensions of $\varepsilon E = \dfrac{[q]}{[L^2]} = \dfrac{[AT]}{[L^2]}$.
Dimensions of $\varepsilon \dfrac{\partial E}{\partial t} = \dfrac{[AT L^{-2}]}{[T]} = [AL^{-2}]$.
LHS = RHS, so (c) is dimensionally correct.
For option (d): $W = \tau\theta$
Dimensions of Work $W = [ML^2T^{-2}]$.
Dimensions of Torque $\tau = [F][r] = [MLT^{-2}][L] = [ML^2T^{-2}]$.
Angle $\theta$ is dimensionless.
LHS = RHS, so (d) is dimensionally correct.
Problem : If velocity $[v]$, time $[T]$ and force $[F]$ are chosen as the base quantities, the dimensions of the mass will be
(a) $[FTv^{-1}]$
(b) $[FTv^{-1}]$
(c) $[FTv^2]$
(d) $[FvT^{-1}]$
[JEE Main 2021, 31 Aug Shift-II]
Solution: Ans. (b)
When the velocity $(v)$, time $(T)$ and force $(F)$ are chosen as base quantities, mass is given by:
$$m \propto v^x T^y F^z \dots (i)$$
Using the dimensional formula of all quantities:
$$[M^1 L^0 T^0] = [LT^{-1}]^x [T]^y [MLT^{-2}]^z$$
$$[M^1 L^0 T^0] = [M^z L^{x+z} T^{-x+y-2z}]$$
Comparing the powers of dimensions on both sides, we get:
$z = 1$; $x + z = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1$; $-x + y – 2z = 0 \Rightarrow -(-1) + y – 2(1) = 0 \Rightarrow 1 + y – 2 = 0 \Rightarrow y = 1$.
Substituting these values in Eq. (i), we get:
$$m \propto v^{-1} T^1 F^1$$
$$m = [FTv^{-1}]$$
Problem : If force $(F)$, length $(L)$ and time $(T)$ are taken as the fundamental quantities. Then what will be the dimension of density?
(a) $[FL^{-4}T^2]$
(b) $[FL^{-3}T^2]$
(c) $[FL^{-5}T^2]$
(d) $[FL^{-3}T^{-3}]$
[JEE Main 2021, 27 Aug Shift-II]
Solution: Ans. (a)
The dimensional formula of density $[D] = [ML^{-3}T^0]$.
Since the dimensional formula of force $[F] = [MLT^{-2}]$, length $[L] = [M^0 L^1 T^0]$, and time $[T] = [M^0 L^0 T^1]$.
$$\therefore [D] = [F^a L^b T^c]$$
$$[ML^{-3}T^0] = [MLT^{-2}]^a [L]^b [T]^c$$
$$[ML^{-3}T^0] = [M^a L^{a+b} T^{-2a+c}]$$
Comparing powers of dimensions on both sides, we get:
$a = 1$; $a + b = -3 \Rightarrow 1 + b = -3 \Rightarrow b = -4$; $-2a + c = 0 \Rightarrow c = 2a \Rightarrow c = 2 \times 1 = 2$.
$\therefore$ The dimensional formula of density will be $[F^1 L^{-4} T^2]$.
Problem : Which of the following is not a dimensionless quantity?
(a) Relative magnetic permeability $(\mu_r)$
(b) Power factor
(c) Permeability of free space $(\mu_0)$
(d) Quality factor
[JEE Main 2021, 27 Aug Shift-I]
Solution: Ans. (c)
Relative magnetic permeability $(\mu_r)$: It is the ratio of permeability of medium to the permeability of free space i.e., $\mu_r = \mu_m / \mu_0$.
As it is a ratio of permeabilities, it is unitless and dimensionless.
Power factor $(\cos \phi)$: It is the cosine of the phase difference between alternating current and alternating voltage. It has only a numerical value, so it is unitless and dimensionless.
Permeability of free space $(\mu_0)$: It is the ratio of magnetising field induction $(B)$ to magnetising field intensity $(H)$ i.e., $\mu_0 = B/H$. Since $H = nI$, $[H] = [L^{-1} A]$. Force on a conductor $F = BIl \Rightarrow B = F/Il$. $$[B] = \frac{[MLT^{-2}]}{[A][L]} = [MT^{-2}A^{-1}]$$ $$\therefore [\mu_0] = \frac{[MT^{-2}A^{-1}]}{[AL^{-1}]} = [MLT^{-2}A^{-2}]$$
Quality factor: It is the ratio of energy stored to the energy dissipated per cycle. As it is a ratio of energies, it is dimensionless.
Hence, permeability of free space is not a dimensionless quantity.
Problem : If $E, L, M$, and $G$ denote the quantities as energy, angular momentum, mass and constant of gravitation respectively,
then the dimension of $P$ in the formula $P = EL^2 M^{-5} G^{-2}$ is
(a) $[M^0 L^1 T^0]$
(b) $[M^1 L^1 T^{-2}]$
(c) $[M^1 L^1 T^{-2}]$
(d) $[M^0 L^0 T^0]$
[JEE Main 2021, 26 Aug Shift-I]
Solution: Ans. (d)
Dimension of energy, $E = [ML^2 T^{-2}]$; Angular momentum, $L = [ML^2 T^{-1}]$; Mass, $M = [M]$; Gravitational constant, $G = [M^{-1} L^3 T^{-2}]$.
The dimension of $P = E L^2 M^{-5} G^{-2}$ is:
$$[P] = \frac{[ML^2 T^{-2}] [ML^2 T^{-1}]^2}{[M]^5 [M^{-1} L^3 T^{-2}]^2}$$
$$[P] = \frac{[ML^2 T^{-2}] [M^2 L^4 T^{-2}]}{[M^5] [M^{-2} L^6 T^{-4}]} = \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^0 L^0 T^0]$$
Problem : The force is given in terms of time $t$ and displacement $x$ by the equation :
$F = A \cos(Bx) + C \sin(Dt)$. The dimensional formula of $AD/B$ is
(a) $[M^0 L^1 T^{-1}]$
(b) $[ML^2 T^{-3}]$
(c) $[M^1 L^1 T^{-2}]$
(d) $[M^2 L^2 T^{-3}]$
[JEE Main 2021, 25 July Shift-II]
Solution Ans. (b) :
Given $F = A \cos(Bx) + C \sin(Dt)$.
$Bx$ must be dimensionless: $[B][L] = [M^0 L^0 T^0] \Rightarrow [B] = [L^{-1}]$.
$Dt$ must be dimensionless: $[D][T] = [M^0 L^0 T^0] \Rightarrow [D] = [T^{-1}]$.
$A$ must have dimensions of force: $[A] = [MLT^{-2}]$. $$[AD/B] = \frac{[MLT^{-2}][T^{-1}]}{[L^{-1}]} = [ML^2 T^{-3}]$$
Problem : If time $(t)$, velocity $(v)$ and angular momentum $(l)$ are taken as the fundamental units, then the dimension of mass $(m)$ in terms of $t, v$ and $l$ is
(a) $[t^{-1} v^{-1} l^2]$
(b) $[t^1 v^2 l^{-1}]$
(c) $[t^{-2} v^{-1} l^1]$
(d) $[t^{-1} v^{-2} l^1]$
[JEE Main 2021, 20 July Shift-II]
Solution: Ans. (d)
Let
$$m = k t^a v^b l^c$$
$$[M^1 L^0 T^0] = [T]^a [LT^{-1}]^b [ML^2 T^{-1}]^c$$
$$[M^1 L^0 T^0] = [M^c L^{b+2c} T^{a-b-c}]$$
Comparing powers: $c = 1$; $b + 2c = 0 \Rightarrow b + 2(1) = 0 \Rightarrow b = -2$; $a – b – c = 0 \Rightarrow a – (-2) – 1 = 0 \Rightarrow a + 1 = 0 \Rightarrow a = -1$.
$$\therefore m \propto t^{-1} v^{-2} l^1$$
Pronlem : The entropy of any system is given by $S = \dfrac{\alpha}{\beta} \ln \left[ \frac{\mu k R}{J \beta^2} + 3 \right]$, where $\alpha$ and $\beta$ are constants; $\mu, J, k$ and $R$ are number of moles, mechanical equivalent of heat, Boltzmann constant and gas constant, respectively. Choose the incorrect option.
(a) $\alpha$ and $J$ have the same dimensions.
(b) $S, k, \beta$ and $\mu R$ have the same dimensions.
(c) $S$ and $\alpha$ have different dimensions.
(d) $\alpha$ and $k$ have the same dimensions.
[JEE Main 2021, 20 July Shift-I]
Solution: Ans. (d)
$S = \Delta Q/T \Rightarrow [S] = [ML^2 T^{-2} K^{-1}]$.
Boltzmann constant $k = \text{Energy}/T \Rightarrow [k] = [ML^2 T^{-2} K^{-1}]$.
Thus $[S] = [k]$.
The term in $\ln$ is dimensionless: $[\mu k R / J \beta^2] = [M^0 L^0 T^0]$.
Mechanical equivalent $[J] = [M^0 L^0 T^0]$.
$$[\beta^2] = [\mu k R] \Rightarrow [\beta] = [ML^2 T^{-2} K^{-1}]$$
From $S = \alpha/\beta$, $[\alpha] = [S][\beta]$.
This results in $\alpha$ and $k$ having different dimensions.
Problem : If $C$ and $V$ represent capacity and voltage respectively, then what are the dimensions of $\lambda$, where $C/V = \lambda$?
(a) $[M^{-2} L^{-3} I^2 T^6]$
(b) $[M^{-3} L^{-4} I^3 T^7]$
(c) $[M^{-1} L^{-3} I^{-2} T^{-7}]$
(d) $[M^{-2} L^{-4} I^3 T^7]$
[JEE Main 2021, 26 Feb Shift-II]
Solution: Ans. (d)
$[C] = [M^{-1} L^{-2} T^4 I^2]$. $[V] = [ML^2 T^{-3} I^{-1}]$.
$$[\lambda] = [C/V] = \dfrac{[M^{-1} L^{-2} T^4 I^2]}{[ML^2 T^{-3} I^{-1}]} = [M^{-2} L^{-4} T^7 I^3]$$
Problem : In a typical combustion engine, the work done by a gas molecule is given by
$W = \alpha^2 \beta e^{-\beta x^2 / kT}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature.
If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be:
(a) $[MLT^{-2}]$
(b) $[M^0LT^0]$
(c) $[M^2LT^{-2}]$
(d) $[MLT^{-1}]$
[2021, 26 Feb Shift-I]
Solution: Ans. (b)
Given the equation:
$$W = \alpha^2 \beta e^{-\beta x^2 / kT}$$
Step 1: Determine the dimensions of $\beta$
The exponent of an exponential function must be dimensionless. Therefore, the term $\dfrac{\beta x^2}{kT}$ is dimensionless ($[M^0L^0T^0]$).
$$[\beta] = \dfrac{[kT]}{[x^2]}$$
We know that the average kinetic energy of a molecule is $\frac{3}{2}kT$, so the dimensions of $kT$ are the same as energy: $[ML^2T^{-2}]$.
$$[\beta] = \dfrac{[ML^2T^{-2}]}{[L^2]} = [MT^{-2}]$$
Step 2: Determine the dimensions of $\alpha$
The term $e^{-\beta x^2 / kT}$ is dimensionless. Therefore, the dimensions of work $W$ must be equal to the dimensions of the product $\alpha^2 \beta$.
$$[W] = [\alpha^2 \beta]$$
We know the dimensions of work $W$ are $[ML^2T^{-2}]$.
$[ML^2T^{-2}] = [\alpha^2] [MT^{-2}]$
Dividing both sides by $[MT^{-2}]$:
$$[\alpha^2] = \frac{[ML^2T^{-2}]}{[MT^{-2}]} = [L^2]$$
Taking the square root:
$$[\alpha] = [L] = [M^0L^1T^0]$$
Thus, the dimensions of $\alpha$ correspond to length.
Problem : If $e$ is electronic charge, $c$ is speed of light and $h$ is Planck’s constant, the quantity $\dfrac{e^2}{4 \pi \epsilon_0 hc}$ has dimensions of
(a) $[MLT^0]$
(b) $[MLT^{-1}]$
(c) $[M^0 L^0 T^0]$
(d) $[LC^{-1}]$
[2021, 25 Feb Shift-I]
Solution: Ans. (c)
Dimensional formula of $[e] = [IT]$, $[h] = [ML^2 T^{-1}]$, and $[c] = [LT^{-1}]$.
$[1/4 \pi \epsilon_0] = [ML^3 T^{-4} I^{-2}]$.
Therefore:
$$\left[ \dfrac{e^2}{4 \pi \epsilon_0 hc} \right] = [M^0 L^0 T^0]$$
Problem : The work done by a gas molecule is $W = \alpha \beta^2 e^{-\dfrac{x^2}{\alpha kT}}$.
Dimensions of $\beta$ will be
(a) $[M^2 L T^2]$
(b) $[M^0 L T^0]$
(c) $[MLT^{-2}]$
(d) $[ML^2 T^{-2}]$
[2021, 24 Feb Shift-I]
Solution: Ans. (c)
Given the equation for work done:
$$W = \alpha \beta^2 e^{-\dfrac{x^2}{\alpha kT}}$$
According to the principle of homogeneity, the exponent of an exponential function must be a dimensionless quantity.
Therefore, the dimensions of $\dfrac{x^2}{\alpha kT}$ are $[M^0 L^0 T^0]$.
1. Finding the Dimensions of $\alpha$
From the exponent:
$$[\alpha] = \frac{[x^2]}{[k][T]}$$
Substituting the known dimensions:
- Displacement $[x] = [L]$
- Boltzmann constant $[k] = [M^1 L^2 T^{-2} K^{-1}]$
- Temperature $[T] = [K]$
$$[\alpha] = \frac{[L^2]}{[M^1 L^2 T^{-2} K^{-1}][K^1]}$$
$$[\alpha] = [M^{-1} T^2]$$
2. Finding the Dimensions of $\beta$
The exponential term $e^{-\dfrac{x^2}{\alpha kT}}$ is dimensionless. Therefore, the dimensions of work ($W$) must be equal to the dimensions of the pre-exponential factor $\alpha \beta^2$.
$$[W] = [\alpha \beta^2]$$
$$[\beta^2] = \frac{[W]}{[\alpha]}$$
Substituting the dimensions of Work $[W] = [M^1 L^2 T^{-2}]$ and our calculated $[\alpha]$:
$$[\beta^2] = \frac{[M^1 L^2 T^{-2}]}{[M^{-1} T^2]}$$
$$[\beta^2] = [M^2 L^2 T^{-4}]$$
Taking the square root to find $[\beta]$:
$$[\beta] = [M^1 L^1 T^{-2}]$$
Correct Option: (c) $[M L T^{-2}]$
Problem : If speed $V$, area $A$ and force $F$ are chosen as fundamental units, then the dimensional formula of Young’s modulus will be
(a) $[FA^2 V^{-3}]$
(b) $[FA^{-1} V^0]$
(c) $[FA^2 V^{-2}]$
(d) $[FA^2 V^{-1}]$
[2020, 2 Sep Shift-I]
Solution: Ans. (b)
Let Let Young’s modulus is related to speed, area and force, as
$$Y = F^x A^y V^z$$
$$[ML^{-1} T^{-2}] = [MLT^{-2}]^x [L^2]^y [LT^{-1}]^z$$
Comparing power of similar quantities,
we have $x = 1, x + 2y + z = − 1, −2x − z = −2$
Solving the powers results in $x=1, y=-1, z=0$.
$$Y = [FA^{-1} V^0]$$
Hence, correct option is (b).
Problem : If momentum $P$, area $A$ and time $T$ are fundamental, dimension of energy is
(a) $[P A T^{-2}]$
(b) $[P A^{-1} T^2]$
(c) $[P A^{1/2} T^{-1}]$
(d) $[P A^{1/2} T^{-1}]$
[2020, 2 Sep Shift-II]
Solution: Ans. (c)
Let Let dimensions of energy E in terms of momentum P, area A and timeT are
$$[E] = [P^x A^y T^z]$$
Substituting dimensions of fundamental quantities for E, P, A and T, we have
$$[ML^2 T^{-2}] = [MLT^{-1}]^x [L^2]^y [T]^z$$
Equating powers of same physical quantities on both sides,
we have x = 1, x + 2y = 2 and −x + z = −2
Solving gives $x=1, y=1/2, z=-1$.
∴Dimensional formula of [E]
$$[E] = [P A^{1/2} T^{-1}]$$
Hence, correct option is (c).
Problem : Amount of solar energy received on the earth’s surface per unit area per unit time is defined as solar constant. Dimensional formula of solar constant is
(a) $[MLT^{-2}]$
(b) $[ML^0T^{-3}]$
(c) $[ML^2T^{-1}]$
(d) $[ML^2T^{-2}]$
[2020, 3 Sep Shift-II]
Solution: Ans. (b)
$$\text{Solar constant} = \frac{\text{Solar energy}}{\text{Area} \times \text{Time}}$$
$$= \frac{[ML^2T^{-2}]}{[L^2][T]}$$
$$= [ML^0T^{-3}]$$
Hence, correct option is (b).
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