Physics M.C.Q (1 Marks)

1. $\lambda>0$

2. 90°

Explanation:

Given $\vec{\text{A}}=\hat{\text{i}}+\hat{\text{j}}$

$\vec{\text{B}}=\hat{\text{i}}-\hat{\text{j}}$

$\vec{\text{A}}.\vec{\text{B}}=|\text{A}||\text{B}|\cos\theta$

$\cos\theta=\frac{\vec{\text{A}}.\vec{\text{B}}}{|\text{A}||\text{B}|}$

$=\frac{(\hat{\text{i}}+\hat{\text{j}}).(\hat{\text{i}}-\hat{\text{j}})}{\sqrt{1^2+1^2}\times\sqrt{1^2+(-1)^2}}=\frac{1-1}{2}=0$

$\Rightarrow\cos\theta=\cos90$

$\therefore\theta=90^\circ.$ Hence, verifies the option (b).

2. $5\sqrt{3}$

3. 100m

Explanation:

projectile is fired at $\theta=15^\circ,\ \text{R} = 50\text{m}$ 

$\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$

$50=\frac{\text{u}^2\sin2\times15^\circ}{\text{g}}\Rightarrow\text{u}^2=50\text{g}\times2$

$\text{u}^2=100\text{g}$

Now $\theta = 45^\circ,\ \text{u}^2=100\text{g}$

$\therefore\ \text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}=\frac{100\text{g}\times\sin2\times45^\circ}{\text{g}}$

$\Rightarrow\ \text{R}=100\text{m}$

So, this verifies option (c).

4. $\frac{2\pi}{(30\sqrt{2})}$

Explanation:

The angle described at the centre by the length of second hand of a watch in 15 second

$=90^\circ=\frac{\pi}{2}$ radians Figure.

Linear speed $\text{v}=\text{r}\omega=\frac{\text{r}\theta}{\text{t}}$

$=\frac{1\times\big(\frac{\pi}{2}\big)}{15}=\frac{\pi}{30}\text{cm s}^{-1}$

Magnitude of change in velocity in 15 sec;

$|\Delta\vec{\text{v}}|=|\vec{\text{v}}_2-\vec{\text{v}}_1|=\sqrt{\text{v}^2_2-\text{v}_1^2}$

$=\sqrt{\text{v}^2+\text{v}^2}=\sqrt{2}\text{v}$

$=\sqrt{2}\frac{\pi}{30}=\frac{2\pi}{30\sqrt{2}}$

2. $5\sqrt{2}(\hat{\text{i}}+\hat{\text{j}})$

2. a, p and b are positive while q is negative.

Main concept used: Sign of a, b, p and q are the sign of their resolving components in the XY direction.

Explanation: Components along X and Y axis of the vector $\vec{\text{u}}$ are both +X and Y direction, so a, b are positive.

Now if we resolve$\vec{\text{v}}$ its X component is in +ve X direction but Y component will be in -ve Y direction. Hence, a, b and p are positive but q is negative. Verifies option (b).

1. $20\text{kmh}^{-1}$

Explanation:

When the man is at rest with respect to the ground, the rain comes to him at an angle 30° with the vertical. This is the direction of the velocity of raindrops with respect to the ground.

Here, V_r,g = Velocity of the rain with respect to the ground

V_m,g = Velocity of the man with respect to the ground

and V_r,m = Velocity of the rain with respect to the man.

Here, V_m,g = 10 kmh^-1

$\text{V}_{\text{r, g}}=\frac{10}{\sin30^\circ}=20\text{kmh}^{-1}$

3. r₁ : r₂

Explanation:

As, centripetal acceleration is given as, $\text{a}_\text{c}=\frac{\text{v}^2}{\text{r}}$

For the first body of mass m₁, $\text{a}_{\text{c}_1}=\frac{\text{v}_1^2}{\text{r}_1}$

For the second body of mass m₂, $\text{a}_{\text{c}_2}=\frac{\text{v}_2^2}{\text{r}_2}$

Also time to complete one revolution by both body is same.

Hence, $\frac{2\pi\text{r}_1}{\text{v}_1}=\frac{2\pi\text{r}_2}{\text{v}_2}$

$\Rightarrow\ \frac{\text{v}_1}{\text{v}_2}=\frac{\text{r}_1}{\text{r}_2}\dots(\text{i})$

i.e., $\text{a}_{\text{c}_1}:\text{a}_{\text{c}_2}=\frac{\text{v}_1^2}{\text{r}_1}\times\frac{\text{r}_2}{\text{v}^2_2}$ [from eq. (i)]

$=\frac{\text{r}_1^2}{\text{r}_2^2}\times\frac{\text{r}_2}{\text{r}_1}=\frac{\text{r}_1}{\text{r}_2}=\text{r}_1:\text{r}_2$​​​​​​​

2. $\frac{\text{r}_\text{A}}{\text{r}_\text{B}}$

3. $\frac{1}{2}$

2. r is along positive X-axis.

Explanation:

Consider a vector $\vec{\text{R}}$ in X-Y plane as shown in figure. If we draw orthogonal vectors $\vec{\text{R}}_{\text{x}}$ and $\vec{\text{R}}_{\text{y}}$ along x and y axes respectively, by law of vector addition, $\vec{\text{R}}=\vec{\text{R}}_{\text{x}}+\vec{\text{R}}_{\text{y}}$

The magnitude of component of r along X-axis

$\text{r}_\text{x}=|\text{r}|\cos\theta$

$(\text{r}_\text{x})_{\text{maximum}}=|\text{r}|(\cos\theta)_{\text{maximum}}$

$\text{r}_\text{x}=|\text{r}|\cos\theta$

$=|\text{r}|\cos0^\circ=|\text{r}|$ $(\because\cos\theta$ is maximum if $\theta=0^\circ)$

As $\theta=0^\circ,$

r is along positive x-axis.

1. $-\text{y v}\hat{\text{k}}$

2. 5ms^-1

Explanation:

At the highest point of the angular projection, the velocity of projectile has only horizontal component velocity $=\text{u}\cos\theta=10\cos60^\circ=5\text{ms}^{-1}.$​​​​​​​

4. $|\text{B}|\neq|\text{A}|$

2. Ranges of A and C are less than that of B.

Explanation:

When a body is projected at an angle with the horizontal with initial velocity u, then the horizontal range R of projectile is $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}.$

Clearly, for maximum horizontal range $\sin2\theta=1$ or $2\theta=90^\circ$ or $\theta=45^\circ.$

Hence, in order to achieve maximum range, the body should be projected at 45°.

In this case $\text{R}_\text{max}=\frac{\text{u}^2}{\text{g}}$

Hence, ranges of A and C are less than that of B.

2. $\sin\theta=\frac{1}{\sqrt{6}}$

Explanation:

Given u = 60ms^-1

$\therefore$ maximum height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$

$\Rightarrow\ 30=\frac{(60)^2\sin^2\theta}{2\text{g}}$

$\Rightarrow\ \sin^2\theta=\frac{30\times2\text{g}}{60\times60}=\frac{10}{60}$

$\Rightarrow\ \sin\theta=\frac{1}{\sqrt{6}}$​​​​​​​

1. $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.

Explanation:

$|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}|$

Applying dot product,

$|\vec{\text{A}}+\vec{\text{B}}|.|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}|.|\vec{\text{A}}|$

$\Rightarrow\ |\vec{\text{A}}|^2+2\vec{\text{A}}.\vec{\text{B}}+|\vec{\text{B}}|^2=|\text{A}|^2$

$\Rightarrow\ |\vec{\text{B}}|^2=-2\vec{\text{A}}.\vec{\text{B}}$

$\Rightarrow\ |\vec{\text{B}}|=0$

Therefore, option (a) is correct.

1. angle of projection: q₁ > q₂
2. time of flight: T₁ > T₂

Explanation:

1. According to the problem,

$\Rightarrow\ \text{H}_1>\text{H}_2$

$\Rightarrow\ \frac{\text{v}^2_0\sin^2\theta_1}{2\text{g}}>\frac{\text{v}^2_0\sin^2\theta_2}{2\text{g}}$

$\Rightarrow\ \sin^2\theta_1>\sin^2\theta_2$

$\Rightarrow\ \sin^2\theta_1-\sin^2\theta_2>0$

$\Rightarrow\ (\sin\theta_1-\sin\theta_2)(\sin\theta_1+\sin\theta_2)>0$

Thus, either $\sin\theta_1+\sin\theta_2>0$

$\Rightarrow\ \sin\theta_1-\sin\theta_2>0$

$\Rightarrow\ \sin\theta_1>\sin\theta_2\text{ or }\theta_1>\theta_2$

Hence option (a) is correct.

2. Time of flight, $\text{T}=\frac{2\text{u}\sin\theta}{\text{g}}=\frac{2\text{v}_0\sin\theta}{\text{g}}$

Thus, $\text{T}_1=\frac{2\text{v}_0\sin\theta_1}{\text{g}}$ and $\text{T}_2=\frac{2\text{v}_0\sin\theta_2}{\text{g}}$

(Here, T₁ = Time of flight of first particle and T₂ = Time of flight of second particle).

As, $\sin\theta_1>\sin\theta_2$

Thus, $\text{T}_1>\text{T}_2$

Hence option (b) is correct.

3. We know that Range, $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}=\frac{\text{v}^2_0\sin2\theta}{\text{g}}$

Range of first particle $=\text{R}_1=\frac{\text{u}^2_0\sin2\theta_1}{\text{g}}$

Range of second particle $=\text{R}_2=\frac{\text{u}^2_0\sin2\theta_2}{\text{g}}$

Given, $\sin\theta_1>\sin\theta_2$

$\Rightarrow\ \sin2\theta_1>\sin2\theta_2$

$\Rightarrow\ \frac{\text{R}_1}{\text{R}_2}=\frac{\sin2\theta_1}{\sin2\theta_2}>1$

$\Rightarrow\ \text{R}_1>\text{R}_2$

But if $\theta_1+\theta_2=90^\circ,$ then $\text{R}_1=\text{R}_2$

Hence option (c) is incorrect.

Important points about time of flight: For complementary angles of projection $\theta\text{ and }90^\circ-\theta.$

1. Ratio of time of flight $=\frac{\text{T}_1}{\text{T}_2}=\frac{2\text{u}\sin\frac{\theta}{\text{g}}}{2\text{u}\sin\frac{(90^\circ-\theta)}{\text{g}}}=\tan\theta$

$\Rightarrow\ \frac{\text{T}_1}{\text{T}_2}=\tan\theta$

2. Multiplication of time of flight $=\text{T}_1\text{T}_2=\frac{2\text{u}\sin\theta}{\text{g}}\frac{2\text{u}\cos\theta}{\text{g}}$

$\Rightarrow\ \text{T}_1\text{T}_2=\frac{2\text{R}}{\text{g}}$

2. $5\sqrt{5}\text{ ms}^{-1}$

1. $\text{v}_\text{av}=\frac{1}2\big[\text{v}(\text{t}_1)+\text{v}(\text{t}_2)\big]$

3. $\text{r}=\frac{1}2(\text{v}(\text{t}_2)-\text{v}(\text{t}_1))(\text{t}_2-\text{t}_1)$

Explanation:

When an object covers a displacement $\Delta\text{r}$ in time $\Delta\text{t},$ its average velocity is given by $\vec{\text{v}}_\text{avg}=\frac{\overrightarrow{\Delta\text{r}}}{\Delta\text{t}}=\frac{\text{r}_2-\text{r}_1}{\text{t}_2-\text{t}_1}$ where r₁ and r₂ are position vectors corresponding to time t₁ and t₂.

If the velocity of an object changes from v₁ to v₂ in time $\Delta\text{t},$ average acceleration is given by

$\text{a}_\text{av}=\frac{\Delta\text{v}}{\Delta\text{t}}=\frac{\text{v}_2-\text{v}_1}{\text{t}_2-\text{t}_1}$

But, when acceleration is non-uniform,

$\text{v}_\text{av}\neq\frac{\text{v}_1+\text{v}_2}{2}$

Option (c) is similar to the relation $\vec{\text{r}}=\frac{1}2\text{at}^2$ which is not correct if initial velocity is given.

So (b) and (d) are the correct relations for the uniform acceleration.

2. $\frac{4}{\sqrt{3}}$

1. $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$ is not zero unless $\vec{\text{B}},\ \vec{\text{C}}$ are parallel.

3. If A, B, C define a plane, $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}$$$ is in that plane.

Explanation:

$\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}}=0$

So $\vec{\text{A}},\ \vec{\text{B}}$ and $\vec{\text{C}}$ are in a plane and can be represented bt the three sides of a triangle taken in order.

1. $\vec{\text{B}}\times(\vec{\text{A}}+\vec{\text{B}}+\vec{\text{C}})=\vec{\text{B}}\times0=0$

$\vec{\text{B}}\times\vec{\text{A}}+\vec{\text{B}}\times\vec{\text{B}}+\vec{\text{B}}\times\vec{\text{C}}=0$

$\vec{\text{B}}\times\vec{\text{A}}+0+\vec{\text{B}}\times\vec{\text{C}}=0$

$\vec{\text{B}}\times\vec{\text{A}}=-\vec{\text{B}}\times\vec{\text{C}}$

$\vec{\text{A}}\times\vec{\text{B}}=\vec{\text{B}}\times\vec{\text{C}}\ \dots(\text{i})$

Or $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=(\vec{\text{B}}\times\vec{\text{C}})\times\vec{\text{C}}$

$\therefore$ It cannot be zero

$\vec{\text{B}}\times\vec{\text{C}}$ will be zero if $\vec{\text{B}}\times\vec{\text{C}}$ are parallel or antiparallel.

i.e, $(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=[\text{BC}\sin0^\circ]\times\vec{\text{C}}$

$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=0$ only if $\vec{\text{B}}||\vec{\text{C}}$

Hence option (a) is verified.

2. $(\vec{\text{A}}\times\vec{\text{B}})=\vec{\text{B}}\times\vec{\text{C}}$ [from (i)]

$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=(\vec{\text{B}}\times\vec{\text{C}}).\vec{\text{C}}$

if $\vec{\text{B}}||\vec{\text{C}}$

$\vec{\text{B}}\times\vec{\text{C}}=\text{BC}\sin0^\circ=0$

$\therefore\ (\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=0\text{ IF }\vec{\text{B}}||\vec{\text{C}}$

So option (b) is not verified.

3. $(\vec{\text{A}}\times\vec{\text{B}})=\vec{\text{X}}$

The direction of x is perpendicular to both planes containing A and B.

$(\vec{\text{A}}\times\vec{\text{B}})\times\vec{\text{C}}=\vec{\text{X}}\times\vec{\text{C}}=\vec{\text{Y}}$

The direction of $\vec{\text{Y}}$ is perpendicular to the plane of $\vec{\text{X}}\text{ and }\vec{\text{C}}$ which again become in the plane of $\vec{\text{A}},\ \vec{\text{B}},\ \vec{\text{C}}$ but perpendicular to the plane of $\vec{\text{X}}\text{ and }\vec{\text{C}}.$ Hence option (c) is also verified.

4. $|\text{A}|^2+|\text{B}|^2=|\text{C}|^2$ given

It shows that angle between $\vec{\text{A}}\text{ and }\vec{\text{B}}$ is 90°

$=|\text{A}||\text{B}||\text{C}|\cos\theta\neq|\text{A}||\text{B}||\text{C}|$

$(\vec{\text{A}}\times\vec{\text{B}}).\vec{\text{C}}=|\text{A}||\text{B}||\text{C}|\cos\theta$

Does not verified option (d).

4. $\frac{\text{A}}{|\text{A}|}$

2. Impulse and area.

Explanation:

We know that impulse J = F. $\Delta\text{t}=\Delta\text{p},$ where F is force, At is time duration and Ap is change in momentum. As $\Delta\text{p}$ is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector direction of area vector is perpendicular to its plane.

3. They have opposite directions.

Explanation:

Multiplying a vector A by a negative number $\lambda$ gives a vector $\lambda\text{A},$ whose directions opposite to the direction of A and it's magnitude is $-\lambda$ times |A|.

2. $\tan^{-1}\Big(\frac{2}{3}\Big)$

Explanation:

As $\vec{\text{A}}=2\hat{\text{i}}+2\hat{\text{j}},$ therefore A_x = 2 and A_y = 3. If $\theta$ is the angle which $\vec{\text{A}}$ encloses with y-axis, then.

$\tan\theta=\frac{\text{A}_\text{x}}{\text{A}_\text{y}}=\frac{2}{3}$ or $\theta=\tan^{-1}\Big(\frac{2}{3}\Big)$​​​​​​​

1. $10(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$

2. $\sqrt{0.11}$

Explanation:

Here, (0.5)² + (0.8)² + (c)² = 1

or $\text{c}=\sqrt{0.11}$