Units and Dimensions
JEE Advanced-2022
Q1
Allen
JEE-Advanced PYQs
Numerical
In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$ , electron mass $m_e$ , Planck's constant $h$ , and Coulomb's constant $k = \frac{1}{4\pi \epsilon_0}$ , where $\epsilon_0$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B] = [e]^{\alpha} [m_e]^{\beta} [h]^{\gamma} [k]^{\delta}$ . The value of $\alpha + \beta + \gamma + \delta$ is ____.
Correct Answer: 4
Explanation:
Ans. (4)
Ans. (4) $B = e^{\alpha}(m_e)^{\beta}h^{\gamma}k^{\delta}$ $[B] = [e^{\alpha}][m_e]^{\beta}[h]^{\gamma}[k^{\delta}]$ $[M^1 T^{-2}A^{-1}] = [AT]^{\alpha}[M]^{\beta}[ML^2 T^{-1}]^{\gamma}[ML^3 A^{-2}T^{-4}]^{\delta}$ $M^1 T^{-2}A^{-1} = M^{\beta + \gamma + \delta}L^{2\gamma + 3\delta}T^{\alpha - \gamma - 4\delta}A^{\alpha - 2\delta}$ Compare: $\beta + \gamma + \delta = 1$ ; $2\gamma + 3\delta = 0$ , $\alpha - \gamma - 4\delta = -2$ , $\alpha - 2\delta = -1$ On solving $\alpha = 3$ , $\beta = 2$ , $\gamma = -3$ , $\delta = 2$ $\alpha + \beta + \gamma + \delta = 4$
Ans. (4) $B = e^{\alpha}(m_e)^{\beta}h^{\gamma}k^{\delta}$ $[B] = [e^{\alpha}][m_e]^{\beta}[h]^{\gamma}[k^{\delta}]$ $[M^1 T^{-2}A^{-1}] = [AT]^{\alpha}[M]^{\beta}[ML^2 T^{-1}]^{\gamma}[ML^3 A^{-2}T^{-4}]^{\delta}$ $M^1 T^{-2}A^{-1} = M^{\beta + \gamma + \delta}L^{2\gamma + 3\delta}T^{\alpha - \gamma - 4\delta}A^{\alpha - 2\delta}$ Compare: $\beta + \gamma + \delta = 1$ ; $2\gamma + 3\delta = 0$ , $\alpha - \gamma - 4\delta = -2$ , $\alpha - 2\delta = -1$ On solving $\alpha = 3$ , $\beta = 2$ , $\gamma = -3$ , $\delta = 2$ $\alpha + \beta + \gamma + \delta = 4$
JEE Main-2021
Q2
Allen
JEE-Main PYQs
Numerical
If $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ , the angle between $\vec{P}$ and $\vec{Q}$ is $\theta$ ( $0^\circ < \theta < 360^\circ$ ). The value of 'Īø' will be ____.
Correct Answer: 180
Explanation:
Ans. (180) $\vec{P} \times \vec{Q} = \vec{Q} \times \vec{P}$ $PQ \sin \theta = -PQ \sin \theta$ $\Rightarrow$ $\sin \theta = 0$ $\Rightarrow$ $\theta = 0, 180, 360\dots\dots$ Given $0^\circ < \theta < 360^\circ$ $\Rightarrow$ $\theta = 180^\circ$
JEE Advanced-2018
Q3
Allen
JEE-Advanced PYQs
Numerical
Two vectors $\vec{A}$ and $\vec{B}$ are defined as $\vec{A} = a\hat{i}$ and $\vec{B} = a(\cos\omega t\hat{i} + \sin\omega t\hat{j})$ , where a is a constant and $\omega = \pi/6$ rad s $^{-1}$ . If $|\vec{A} + \vec{B}| = \sqrt{3}|\vec{A} - \vec{B}|$ at time $t = \tau$ for the first time, the value of $\tau$ , in seconds, is ____.
Correct Answer: 2
Explanation:
Ans. (2 [1.99, 2.01])
$\left| \vec {A} + \vec {B} \right| = 2 a \cos {\frac {\omega t}{2}}$
$\left| \vec {A} - \vec {B} \right| = 2 a \sin {\frac {\omega t}{2}}$
So, $2a\cos \frac{\omega t}{2} = \sqrt{3}\left(2a\sin \frac{\omega t}{2}\right)$
$\tan {\frac {\omega t}{2}} = \frac {1}{\sqrt {3}}$
${\frac {\omega t}{2}} = {\frac {\pi}{6}}, {\frac {7 \pi}{6}}, {\frac {1 3 \pi}{6}} \dots$
$t = 2, 1 4, 2 6 \dots \qquad \left(\because \omega = \frac {\pi}{6}\right)$
So, the given condition occurs for the first time at t = 2 seconds.
$\left| \vec {A} + \vec {B} \right| = 2 a \cos {\frac {\omega t}{2}}$
$\left| \vec {A} - \vec {B} \right| = 2 a \sin {\frac {\omega t}{2}}$
So, $2a\cos \frac{\omega t}{2} = \sqrt{3}\left(2a\sin \frac{\omega t}{2}\right)$
$\tan {\frac {\omega t}{2}} = \frac {1}{\sqrt {3}}$
${\frac {\omega t}{2}} = {\frac {\pi}{6}}, {\frac {7 \pi}{6}}, {\frac {1 3 \pi}{6}} \dots$
$t = 2, 1 4, 2 6 \dots \qquad \left(\because \omega = \frac {\pi}{6}\right)$
So, the given condition occurs for the first time at t = 2 seconds.
JEE Advanced-2014
Q4
Allen
JEE-Advanced PYQs
Numerical
To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to $S^{1/n}$ . The value of n is.
Correct Answer: 3
Explanation:
Ans. (3)
$[ d ] = (S) ^ {x} (\rho) ^ {y} (f) ^ {z}$
$L = (M ^ {1} L ^ {0} T ^ {- 3}) ^ {x} (M ^ {1} L ^ {- 3}) ^ {y} (T ^ {- 1}) ^ {z}$
$L = M ^ {x + y} L ^ {- 3 y} T ^ {- 3 x - z}$
$- 3 y = 1 \qquad x + y = 0$
3 x + z = 0
$y = - \frac {1}{3} x - \frac {1}{3} = 0$
z = - 3 x
$x = \frac {1}{3}$
z = - 1
$L = (S)^{1 / 3}(\rho)^{-1 / 3}(f)^{-1}$
n = 3
$[ d ] = (S) ^ {x} (\rho) ^ {y} (f) ^ {z}$
$L = (M ^ {1} L ^ {0} T ^ {- 3}) ^ {x} (M ^ {1} L ^ {- 3}) ^ {y} (T ^ {- 1}) ^ {z}$
$L = M ^ {x + y} L ^ {- 3 y} T ^ {- 3 x - z}$
$- 3 y = 1 \qquad x + y = 0$
3 x + z = 0
$y = - \frac {1}{3} x - \frac {1}{3} = 0$
z = - 3 x
$x = \frac {1}{3}$
z = - 1
$L = (S)^{1 / 3}(\rho)^{-1 / 3}(f)^{-1}$
n = 3
Q5
Allen
NAT
Numerical
If mass is expressed as $v^{x} d^{y} a^{z}$ where v is velocity; d is density and a is acceleration then the value of $x + y + z$ is
Correct Answer: 4
Explanation:
Ans. (4) $m=Rv^{x}dy a^{z}$ (where k is dimensionless) $[M]=[LT^{-1}]^{x}[ML^{-3}]^{y}[LT^{-2}]^{z}$ Solve for x, y and z.
Q6
Allen
NAT
Numerical
The angle subtended by the moon's diameter at a point on the earth is about $0.50^{\circ}$ . Use this and the fact that the moon is about 384000 km away to find the approximate diameter of the moon (in km).
Correct Answer: 3350
Explanation:
Ans. (3350)
$D \approx r _ {m} \theta = (3 8 4 0 0 0) \left(\frac {0 . 5}{1 8 0 / \pi}\right) = 3 3 5 0 k m$
$D \approx r _ {m} \theta = (3 8 4 0 0 0) \left(\frac {0 . 5}{1 8 0 / \pi}\right) = 3 3 5 0 k m$
Q7
Allen
NAT
Numerical
Three particles P, Q and R are moving along the vectors $\vec{A} = \hat{i} + \hat{j}, \vec{B} = \hat{j} + \hat{k}$ and $\vec{C} = -\hat{i} + \hat{j}$ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $\vec{A}$ and $\vec{B}$ . Similarly, particle Q is moving normal to the plane which contains vector $\vec{A}$ and $\vec{C}$ . The angle between the direction of motion of P and Q is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$ . Then the value of x is ____.
Correct Answer: 3
Explanation:
Ans. (3)
Direction of $P$ , $\hat{v}_1 = \pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$
Direction of $Q$ , $\hat{v}_2 = \pm \frac{\vec{A} \times \vec{C}}{|\vec{A} \times \vec{C}|} = \pm \frac{2\hat{k}}{2} = \pm \hat{k}$
Angle between $\hat{v}_{1}$ and $\hat{v}_{2}$ -
$\cos \theta = \frac {\hat {v} _ {1} . \hat {v} _ {2}}{| \hat {v} _ {1} | | \hat {v} _ {2} |} = \frac {\pm 1 / \sqrt {3}}{(1) (1)} = \pm \frac {1}{\sqrt {3}}$
$\Rightarrow x = 3$
Direction of $P$ , $\hat{v}_1 = \pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$
Direction of $Q$ , $\hat{v}_2 = \pm \frac{\vec{A} \times \vec{C}}{|\vec{A} \times \vec{C}|} = \pm \frac{2\hat{k}}{2} = \pm \hat{k}$
Angle between $\hat{v}_{1}$ and $\hat{v}_{2}$ -
$\cos \theta = \frac {\hat {v} _ {1} . \hat {v} _ {2}}{| \hat {v} _ {1} | | \hat {v} _ {2} |} = \frac {\pm 1 / \sqrt {3}}{(1) (1)} = \pm \frac {1}{\sqrt {3}}$
$\Rightarrow x = 3$
Q8
Allen
NAT
Numerical
During a war between Ra1 and G1, the power shot should be by G1 reaches to Ra1. It was found that the power of the power shot fired depends on mass $(m_{0})$ of G1, velocity $(v_{0})$ of his hand and time lag $(t_{0})$ between his thought of firing & actual time when shot was fired. If power of power shot depends on $k^{th}$ power of velocity of his hand, fill k in OMR sheet.
Correct Answer: 2
Explanation:
Ans. (2)
$\left. \begin{array}{c} p \propto m ^ {a} \\ p \propto v ^ {b} \\ p \propto t _ {0} ^ {c} \end{array} \right]$
$p = k [ M ] ^ {a} \left[ L T ^ {- 1} \right] ^ {b} \left[ T \right] ^ {c}$
$M L ^ {2} T ^ {- 3} = M ^ {a} L ^ {b} T ^ {- b + c}$
a = 1
b = 2
- b + c = - 3
- 2 + c = - 3
c = - 1
$p \propto v ^ {2} \Rightarrow p = k v ^ {2}$
$\left. \begin{array}{c} p \propto m ^ {a} \\ p \propto v ^ {b} \\ p \propto t _ {0} ^ {c} \end{array} \right]$
$p = k [ M ] ^ {a} \left[ L T ^ {- 1} \right] ^ {b} \left[ T \right] ^ {c}$
$M L ^ {2} T ^ {- 3} = M ^ {a} L ^ {b} T ^ {- b + c}$
a = 1
b = 2
- b + c = - 3
- 2 + c = - 3
c = - 1
$p \propto v ^ {2} \Rightarrow p = k v ^ {2}$
Q9
Allen
NAT
Numerical
If the resultant of two forces of magnitudes P and Q acting at a point at an angle of $60^{\circ}$ is $\sqrt{7}$ Q, then P/Q is
Correct Answer: 2
Explanation:
Ans. (2)
$R=\sqrt{P^2+Q^2+2PQ\cos\theta}$
$\sqrt{7}Q=\sqrt{P^2+Q^2+2PQ\cos60^\circ}$
$7Q^2=P^2+Q^2+PQ$
$P^2-6Q^2+PQ=0$
$P^2+PQ-6Q^2=0$
$P=\frac{-Q\pm\sqrt{Q^2+24Q^2}}{2}$
$P=\frac{-Q+5Q}{2}=2Q$
$\frac{P}{Q}=2$
$R=\sqrt{P^2+Q^2+2PQ\cos\theta}$
$\sqrt{7}Q=\sqrt{P^2+Q^2+2PQ\cos60^\circ}$
$7Q^2=P^2+Q^2+PQ$
$P^2-6Q^2+PQ=0$
$P^2+PQ-6Q^2=0$
$P=\frac{-Q\pm\sqrt{Q^2+24Q^2}}{2}$
$P=\frac{-Q+5Q}{2}=2Q$
$\frac{P}{Q}=2$
Q10
Allen
NAT
Numerical
Two forces $\vec{F}_1$ and $\vec{F}_2$ of magnitude $10\sqrt{10} N$ each are inclined at an angle of $1.8^{\circ}$ to each other. What is the magnitude (in $N$ ) of vector $\vec{F}_1 - \vec{F}_2$ ? (Take $\pi^2 = 10$ .)
Correct Answer: 1
Explanation:
Ans. (1)
$|\vec{F}_1-\vec{F}_2|=\sqrt{F_1^2+F_2^2-2F_1F_2\cos(1.8^\circ)}$
$1.8^\circ=\frac{1.8}{180}\pi\text{ rad.}=\frac{\pi}{100}\text{ rad}$
$\therefore\ |\vec{F}_1-\vec{F}_2|=\sqrt{1000+1000-2000\cos\frac{\pi}{100}}$
$=\sqrt{2000\left(1-\cos\frac{\pi}{100}\right)}=\sqrt{2000\times2\sin^2\frac{\pi}{200}}$
$=\sqrt{4000\left(\frac{\pi^2}{40000}\right)}=1$
$|\vec{F}_1-\vec{F}_2|=\sqrt{F_1^2+F_2^2-2F_1F_2\cos(1.8^\circ)}$
$1.8^\circ=\frac{1.8}{180}\pi\text{ rad.}=\frac{\pi}{100}\text{ rad}$
$\therefore\ |\vec{F}_1-\vec{F}_2|=\sqrt{1000+1000-2000\cos\frac{\pi}{100}}$
$=\sqrt{2000\left(1-\cos\frac{\pi}{100}\right)}=\sqrt{2000\times2\sin^2\frac{\pi}{200}}$
$=\sqrt{4000\left(\frac{\pi^2}{40000}\right)}=1$
Q11
Allen
NAT
Numerical
If $\hat{i} + 2\hat{j} - n\hat{k}$ is perpendicular to $4\hat{i} + 2\hat{j} + 2\hat{k}$ , then the value of n is
Correct Answer: 4
Explanation:
Ans. (4)
$(\hat{i}+2\hat{j}-n\hat{k})\cdot(4\hat{i}+2\hat{j}+2\hat{k})=0$
$\Rightarrow 4+4-2n=0$
$\Rightarrow n=4$
$(\hat{i}+2\hat{j}-n\hat{k})\cdot(4\hat{i}+2\hat{j}+2\hat{k})=0$
$\Rightarrow 4+4-2n=0$
$\Rightarrow n=4$
Q12
Allen
NAT
Numerical
A force $\vec{F}=5\hat{i}+2\hat{j}+\hat{k}$ displaces a body from a point of coordinate $(1,1,1)$ to another point of coordinates $(2,0,3)$ . Calculate the work done (in J) by the force.
Correct Answer: 5
Explanation:
Ans. (5)
Displacement vector $(\vec{d})=(2-1)\hat{i}+(0-1)\hat{j}+(3-1)\hat{k}$
$\Rightarrow \vec{d}=\hat{i}-\hat{j}+2\hat{k}$
$\vec{F}=5\hat{i}+2\hat{j}+\hat{k}$
Work done $=\vec{F}\cdot\vec{d}=(5\hat{i}+2\hat{j}+\hat{k})\cdot(\hat{i}-\hat{j}+2\hat{k})$
$=5-2+2=5J$
Displacement vector $(\vec{d})=(2-1)\hat{i}+(0-1)\hat{j}+(3-1)\hat{k}$
$\Rightarrow \vec{d}=\hat{i}-\hat{j}+2\hat{k}$
$\vec{F}=5\hat{i}+2\hat{j}+\hat{k}$
Work done $=\vec{F}\cdot\vec{d}=(5\hat{i}+2\hat{j}+\hat{k})\cdot(\hat{i}-\hat{j}+2\hat{k})$
$=5-2+2=5J$
Q13
Allen
NAT
Numerical
Three vectors $\vec{P},\vec{Q}$ and $\vec{R}$ are such that $|\vec{P}| = |\vec{Q} |,\left|\vec{R}\right| = \sqrt{2} |\vec{P} |$ and $\vec{P} +\vec{Q} +\vec{R} = \vec{0}$ . If the angle between $\vec{P}$ and $\vec{R}$ is $\frac{a\pi}{4}$ (in radians) then find the value of $(\alpha)$ .
Correct Answer: 3
Explanation:
Ans. (3)
Q14
Allen
NAT
Numerical
The sum of two force acting at a point is 16N. If their resultant is normal to the smaller force and has a magnitude of 8N, then find the value of smaller force.
Correct Answer: 6
Explanation:
Ans. (6)
Given
$\vec{F}=\vec{F}_1+\vec{F}_2$
$F_1+F_2=16$
$F=8$
$F^2=F_2^2-F_1^2$
$\Rightarrow 8^2=(F_2+F_1)(F_2-F_1)$
$\Rightarrow 64=16(F_2-F_1)$
$\Rightarrow F_2-F_1=4$
$\Rightarrow 16-F_1-F_1=4$
$\Rightarrow F_1=6,N$
Given
$\vec{F}=\vec{F}_1+\vec{F}_2$
$F_1+F_2=16$
$F=8$
$F^2=F_2^2-F_1^2$
$\Rightarrow 8^2=(F_2+F_1)(F_2-F_1)$
$\Rightarrow 64=16(F_2-F_1)$
$\Rightarrow F_2-F_1=4$
$\Rightarrow 16-F_1-F_1=4$
$\Rightarrow F_1=6,N$