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Physics 5 Marks Questions

Model Paper 3 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 6 questions.

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6 questions · timed · auto-graded

Question 15 Marks
A metre scale AB is held vertically with its one end A on the floor and is then allowed to fall. Find the speed of the other end B when it strikes the floor, assuming that the end on the floor does not slip.
Answer
Let $M$ be the mass and Lbe the length of the metre scale. When the upper end of the rod strikes the floor, its centre of gravity falls through height $\frac{L}{2}$.
Image
M.I. of the scale about the lower end $A, I=$ M.I. of the scale about the parallel axis through $C G+M d_2$
$
=I_0+M d^2=\frac{M L^2}{12}+\frac{M L^2}{4}=\frac{M L^2}{3} \quad\left[\because d=\frac{L}{2}\right]
$
Also, $\omega=\frac{v}{r}=\frac{v}{L}$
Gain in rotational K.E.
$
=\frac{1}{2} I \omega^2=\frac{1}{2} \cdot \frac{M L^2}{3} \cdot \frac{v^2}{L^2}=\frac{M v^2}{6}
$
Now, Gain in rotational K.E. $=$ Loss in P.E.
$
\frac{M v^2}{6}=M g \cdot \frac{L}{2} \quad \text { or } \quad v^2=3 g l
$
or $v=\sqrt{3 g L}=\sqrt{3 \times 9.8 \times 1}=5.4 ms^{-1}$
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Question 25 Marks
Derive an expression for moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane.
Answer
Consider a disc of mass $M$ and radius $R$. This disc is made up of many infinitesimally small rings, as shown in the figure. Consider one such ring of mass $( dm )$ and thickness $( dr )$ and radius $( r )$. The moment of inertia $( dl )$ of this small ring is, $dI =$ $( dm ) R ^2$
Image
$\begin{aligned} & I=\int_R^R d I \\ & I =\int_0^R \frac{2 M}{R^2} r^3 d r=\frac{2 M}{R^2} \int_0^R r^3 d r \\ & I =\frac{2 M}{R^2}\left[\frac{r^4}{4}\right]_0^R=\frac{2 M}{R^2}\left[\frac{R^4}{4}-0\right] \\ & I =\frac{1}{2} M R^2\end{aligned}$
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Question 35 Marks
$\hat{i}$ and $\hat{j}$ are unit vectors along x and y -axes respectively. What is the magnitude and direction of vectors $\hat{i}+\hat{j}$ and $\hat{i}$ $\hat{j}$ ? What are the components of a vector $A =2 \hat{ i }+3 \hat{ j }$ along the direction $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$ ?
Answer
i. As we know $\hat{i}$ and $\hat{j}$ are unit vectors, Magnitude of $(\hat{i}+\hat{j})=\sqrt{(1)^2+(1)^2}=\sqrt{2}$ units
If vector $(\hat{ i }+\hat{ j })$ makes an angle of $\theta$ with the $x -$ axis, then $\tan \theta=\frac{A_y}{A_x}=\frac{1}{1}=1=\tan 45^{\circ}$ or $\theta=45^{\circ}$
Image
ii. Similarly, magnitude of
$
(\hat{ i }-\hat{ j })=\sqrt{(1)^2+(-1)^2}=\sqrt{2}
$
If vector $(\hat{ i }-\hat{ j })$ makes an angle $\theta$, with x - axis, then
$
\begin{aligned}
& \tan \theta=\frac{A_y}{A_x}=\frac{(-1)}{1}=-1 \\
& =-\tan 45^{\circ} \Rightarrow \theta=-45^{\circ} \text { with } \hat{i}
\end{aligned}
$
Hence, resultant vector $(\hat{ i }-\hat{ j })$ makes an angle of $45^{\circ}$ from $x$-axis in negative direction.
iii. To determine the component of $A =2 \hat{ i }+3 \hat{ j }$ in the direction of $(\hat{ i }+\hat{ j })$
Let us assume $B=(\hat{ i }+\hat{ j })$, then
$A \cdot B = AB \cos \theta=( A \cos \theta) . B$
or $A \cos \theta=\frac{ A \cdot B }{B}$
$\Rightarrow A \cos \theta=\frac{ A \cdot B }{B}=\frac{(2 \hat{ i }+3 \hat{ j }) \cdot(\hat{ i }+\hat{ j })}{\sqrt{(1)^2+(1)^2}}$
$=\frac{2 \hat{ i } \hat{ i } \cdot 3 \hat{ j } \hat{ j }}{\sqrt{2}}$
$=\frac{2+3}{\sqrt{2}}=\frac{5}{\sqrt{2}}$. This is the component of vector A in the direction of $(\hat{ i }+\hat{ j })$
iv. Unit vector along $(\hat{ i }+\hat{ j }), \hat{ n }=\frac{(\hat{ i }+\hat{ j })}{|\hat{ i }+\hat{ j }|}=\frac{(\hat{ i }+\hat{ j })}{\sqrt{2}}$
Component of A along $(\hat{ i }-\hat{ j })$
The magnitude of the component of A in the direction of $(\hat{ i }-\hat{ j })=\frac{(2 \hat{ i }+3 \hat{ j }) \cdot(\hat{ i }-\hat{ j })}{|\hat{ i }-\hat{ j }|}=\frac{2 \hat{ i } \cdot \hat{ i }-3 \hat{ j } \hat{ j }}{\sqrt{(1)^2+(-1)^2}}=\frac{2-3}{\sqrt{2}}=\frac{-1}{\sqrt{2}}$. This is the component of vector A in the direction of $(\hat{ i }-\hat{ j })$.
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Question 45 Marks
What is meant by resolution of a vector? Prove that a vector can be resolved along two given directions in one and only one way.
Answer
Resolution of a vector. It is the process of splitting a vector into two or more vectors in such a way that their combined effect is same as that of the given vector. The vectors into which the given vector is splitted are called component vectors. A component of a vector in any direction gives a measure of the effect of the given vector in that direction. The resolution of a vector is just opposite to the process of vector addition.
Resolution of a vector along two given directions.
Suppose we wish to resolve a vector $\vec{R}$ in the direction of two coplanar and non-parallel vectors $\vec{A}$ and $\vec{B}$, as shown in Figure.
Image
Suppose $\overrightarrow{O Q}$ represent vector $\vec{R}$ in the directions of $\vec{A}$ and $\vec{B}$.
Q draw lines parallel to vectors $\vec{A}$ and $\vec{B}$ respectively to meet at point P. From triangle law of vector addition.
$
\overrightarrow{O Q}=\overrightarrow{O P}+\overrightarrow{P Q}
$
As $\overrightarrow{O P} \| \vec{A}$ therefore, $\overrightarrow{O P}=\lambda \vec{A}$
As $\overrightarrow{P Q} \| \vec{B}$ therefore, $\overrightarrow{P Q}=\mu \vec{B}$
Here $\lambda$ and $\mu$ are scalar. Hence
$
\vec{R}=\lambda \vec{A}+\mu \vec{B} \ldots \text { (i) }
$
Thus the vector $\vec{R}$ has been resolved in the direction of $\vec{A}$ and $\vec{B}$. Here $\lambda \vec{A}$ is the component of $\vec{R}$ in the direction $\vec{A}$ and $\mu \vec{B}$ is the component in the direction of $\vec{B}$.
Uniqueness of resolution. Let us assume that $\vec{R}$ can be resolved in the directions of $\vec{A}$ and $\vec{B}$ in another way.
Then $\vec{R}=\lambda^{\prime} \vec{A}+\mu^{\prime} \vec{B} \ldots$... (ii)
From equation (i) and (ii), we have
$
\begin{aligned}
& \lambda \vec{A}+\mu \vec{B}=\lambda^{\prime} \vec{A}+\mu^{\prime} \vec{B} \\
& \text { or }\left(\lambda-\lambda^{\prime}\right) \vec{A}=\left(\mu^{\prime}-\mu\right) \vec{B}
\end{aligned}
$
But $\vec{A}$ and $\vec{B}$ are non-zero vectors acting along different directions. The above equation is possible only if
$
\begin{aligned}
& \lambda-\lambda^{\prime}=0 \text { and } \mu^{\prime}-\mu=0 \\
& \text { or } \lambda^{\prime}=\lambda \text { and } \mu^{\prime}=\mu
\end{aligned}
$
Hence there is one and only one way in which a vector $\vec{R}$ can be resolved in the directions of vectors $\vec{A}$ and $\vec{B}$.
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Question 55 Marks
Two identical springs, each of spring factor k, may be connected in the following ways. Deduce the spring factor of the oscillation of the body in each case.
Answer
For each spring,
$
F=-ky \text {...(i) }
$
where $F =$ restoring force, $k =$ spring factor and $y =$ displacement of the spring.
i. In Figure, let the mass m produce a displacement y in each spring and F be the restoring force in each spring. If $k _1$ be the spring factor of the combined system, then
Image
$
2 F=-ky
$
or $F =-\frac{k_1}{2} y$$\ldots(ii)$
Comparing (i) and (ii), we get
$
\frac{k_1}{2}=k \text { or } k_1=2 k
$
ii. In Figure, as the length of the spring is doubled, the mass m will produce double the displacement ( 2 y ). If $k _2$ be the spring factor of the combined system, then
Image
$
F=-k_2(2 y)=-2 k_2 y \ldots(iii)
$
Comparing (i) and (iii),
$
2 k_2=k \text { or } k_2=\frac{k}{2}
$
iii. In figure, the mass m stretches the upper spring and compresses the lower spring, each giving rise to a restoring force F in the same direction. If $k_3$ be the spring factor of the combined system, then
Image
$
\begin{aligned}
& 2 F=-k_3 y \\
& \text { or } F=-\frac{k_3}{2} y \ldots \text { (iv) }
\end{aligned}
$
Comparing (i) and (iv),
$
\frac{k_3}{2}=k \text { or } k_3=2 k
$
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Question 65 Marks
An air chamber of volume $V$ has a neck of the area of cross-section $A$ into which a ball of mass $m$ can move without friction. Show that when the ball is pressed down through some distance and released, the ball executes SHM. Obtain the formula for the time period of this SHM, assuming pressure-volume variations of the air to be
i. isothermal and
ii. adiabatic.
Answer
Oscillations of a ball in the neck of an air chamber. The figure shows an air chamber of volume V , having a neck of area of crosssection $A$ and a ball of mass $m$ fitting smoothly in the neck. If the ball be pressed down a little and released, it starts oscillating up and down about the equilibrium position.
Image
If the ball be depressed by distance $y$, then the decrease in volume of air in the chamber is $\Delta V= Ay$.
$\therefore$ Volume strain $=\frac{\Delta V}{V}=\frac{A y}{V}$
If pressure P is applied to the ball, then hydrostatic stress $= P$
$\therefore$ Bulk modulus of elasticity of air,
$
E=-\frac{P}{\Delta V / V}=-\frac{P}{A y / V} \text { or } P=-\frac{E A}{V} y
$
Restoring force, $F = PA =-\frac{E A y}{V} A=-\frac{E A^2}{V} y$
Thus F is proportional to y and acts in its opposite direction. Hence the ball executes SHM with force constant, $k =\frac{E A^2}{V}$
The period of oscillation of the ball is
$
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{m}{E A^2 / V}}=2 \pi \sqrt{\frac{m V}{E A^2}}
$
i. If the $P - V$ variations are isothermal, then $E = P$,
$
\therefore T=2 \pi \sqrt{\frac{m V}{P A^2}}
$
ii. If the $P - V$ variations are isothermal, then $E =\gamma P$,
$
\therefore T=2 \pi \sqrt{\frac{m V}{\gamma P A^2}}
$
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5 Marks Questions - Physics STD 11 Science Questions - Vidyadip