Grade 11 NEB Practice Set II with solutions | Physics | Physics in Depth

Physics Practice Set II

1. A hose A ball A of mass 0.2 kg, moving with a velocity of 8 m/s, collides directly with a ball B of mass 0.3 kg at rest. Calculate their common velocity if both balls move together. If A had rebounded with a velocity of 3 m/s in the opposite direction after collision, what should be the new velocity of B?
Solution :
For ball 'A',
mass , \(m_1\) = 0.2 kg
velocity before collision, \(u_1\) = 8 m/s

For ball 'B',
mass, \(m_2\) = 0.3 kg
velocity before collision, \(u_2\) = 0

Let \(v\) be the common velocity of both balls after collision.

From the conservation of linear momentum,
\[\begin{align} m_1u_1+m_2u_2&=(m_1+m_2)v\\ 0.2 \times 8 + 0 & = (0.2 + 0.3) v\\ v& = \frac {1.6}{0.5}\\ \therefore v&= 3.2 \hspace{0.1cm} \text{m/s} \end{align}\]
If A had rebounded with a velocity of 3 m/s , then,
velocity of A, \(v_1\) = -3 m/s
velocity of B, \(v_2\) = ?

\[\begin{align} m_1u_1+m_2u_2&=m_1v_1 + m_2v_2\\ 1.6 & = 0.2 \times (-3) + 0.3 \times v_2\\ 1.6 & = -0.6 + 0.3v_2\\ v_2&=\frac{2.2}{0.3}\\ \therefore v_2 & = 7.33 \hspace{0.1cm} \text{m/s} \end{align}\]
Hence, the required solutions are 3.2 m/s and 7.33 m/s respectively.


2. A hose directs a horizontal jet of water moving with a velocity 20 m/s, on a vertical wall. The cross sectional area of the jet is 5 x 10\(^{-4}\)m². If the density of water is 1000 kg/m³. Calculate the force on the wall assuming the water is brought to rest.
Solution :
velocity of water jet before collision, \(u\) = 20 m/s
cross - sectional area of jet, \(A\) = 5 x 10\(^{-4}\)m²
density of water, \(\rho\) = 1000 kg/m³
velocity of water jet after collision, \(v\) = 0
force exerted on the wall by the jet, \(F\) = ?

Now, using the relations, \[F=\frac{m(v-u)}{t}\] \[m=\rho \times V\] and, \[V=A \times l\] We obtain, \[\begin{align} F&=\frac{\rho \times A \times l \times (v-u)}{t}\\ &=\rho \times A \times u \times (v-u) \hspace{0.1cm} \because \frac{l}{t}=u\\ &= 1000 \times 5 \times 10{^{-4}} \times 20 \times (0-20)\\ \therefore F &=-200 \hspace{0.1cm} \text{N} \end{align}\]
Hence, the required solution is 200 N.


3. A train of mass 2x10⁵ kg moves at a constant speed of 72 km/h up a straight inclined against a frictional force of 1.26x10⁴ N. The incline is such that the train rises vertically 1.0 m for every 100 m travelled along the incline. Calculate the necessary power developed by the train.

Solution :
Here, mass of train, m = \(2\times 10^5\) kg, speed, v = 72 km/h, frictional force, f = \(1.26\times 10^4\) N, \(sin\theta=\frac{1}{100}\), Power, P = ?


Since the train is moving straight incline up, \[\begin{align*} \text{Net force}&=F-(f+mgsin\theta)\\ 0&=F-(f+mgsin\theta) \hspace{0.1cm} \because v=\text{const.} \implies a=0\\ F&=f+mgsin\theta\\ &=1.26\times 10^4+2\times 10^5\times 10\times \frac{1}{100}\\ \therefore F&=3.26\times 10^4 \end{align*}\] So, the power developed by the train is, \[\begin{align*} P&=Fv\\ &=3.26\times 10^4 \times \frac{72\times 1000}{3600}\\ \therefore P&=652000 \hspace{0.1cm} \text{Watt} \end{align*}\]
Hence, the required solution is 652000 W .


4. A boat is heading east at 4m/s in a river flowing south at 1m/s. What is the velocity of the boat relative to the earth?
Solution :
In the figure, let \(\vec{v_B}\) denotes the velocity of boat, \(\vec{v_R}\) denotes the velocity of river and \(\vec{v_{BE}}\) be the velocity of boat relative to te earth.


\(v_B\) = 4 m/s
\(v_R\) = 1 m/s
\(v_{BE}\) = ?

Using pythagoras theorem, \[\begin{align} v_{BE}^2 & = v_B ^2 + v_R ^2\\ & = 16 + 1\\ \therefore v_{BE} & = \sqrt{17} = 4.12 \hspace{0.1cm} \text{m/s} \end{align}\]
Now, \[\begin{align} tan\theta&=\frac{v_R}{v_B}\\ \theta&=tan^{-1}(\frac{1}{4}\\ \therefore \theta & = 14.03 ^\circ \end{align}\]
Hence, the velocity of boat relative to earth is 4.12 m/s, 14.03 \(^\circ\) south of east.



5. The element of an electric fire with an output 1kW ,is a cylinder 25cm long and 1.5 cm in diameter .Calculate its temperature when in use, if it behaves as a black body.(the Stefan Constant is 5.7\(\times 10^{-8}\) Wm\(^{-2}\) K\(^{-4}\) .
Solution : Click here ! and scroll down to q.no. (24) for the solution.


6. Find the result of mixing 20 gm of water at 80°C to 40 gm of ice at -10°C.
Solution : Click here ! and scroll down to q.no.(5) of numerical problem solution to find the solution to this question.


7. A copper vessel with a volume of exactly 1.6 m³ at a temperature 20⁰C is filled with Glycerin. If the temperature rises to 30⁰C, how much glycerin will spill out?
Solution : Click here ! and scroll down to q.no.(3) of numerical problem solution to find the solution.


8. Calculate the heat required to convert 20 gm of ice at – 20 \(^\circ\) C to steam at 100\(^\circ\) C. [latent heat of fusion of ice 80 cal/g, latent heat of vaporization 540 cal/g, specific heat capacity of ice 0.5 cal/g\(^\circ\)C]

Solution :




mass of ice, \(m_i\) = 20 gm
latent heat of fusion of ice, \(L_i\) = 80 cal/g
latent heat of vaporization , \(L_v\) = 540 cal/g
specfic heat capacity of ice, \(s_i\) = 0.5 cal/g\(^\circ\)C
required amount of heat = ?
( Just see the figure and go with the flow, dear students!).

Amount of heat required to convert ice at – 20 \(^\circ\) C to ice at 0 \(^\circ\) C is, \[\begin{align} &=m_is_i d\theta\\ &=20 \times 0.5 \times (0-(-20))\\ &=20 \times 0.5 \times 20\\ &=200 \hspace{0.1cm} \text{Cal} \end{align}\]
Amount of heat required to convert ice at 0 \(^\circ\) C to water at 0 \(^\circ\) C is, \[\begin{align} &=m_iL_i\\ &=20 \times 80 \\ &=1600 \hspace{0.1cm}\text{Cal} \end{align}\]
Amount of heat required to convert water at 0 \(^\circ\) C to water at 100 \(^\circ\) C is, \[\begin{align} &=m_ws_w d\theta\\ &=20 \times 1 \times (100-0)\\ &=20 \times 1 \times 100\\ &=2000 \hspace{0.1cm} \text{Cal} \end{align}\]
Amount of heat required to convert water at 100 \(^\circ\) C to steam at 100 \(^\circ\) C is, \[\begin{align} &=m_wL_v\\ &=20 \times 540 \\ &=10800 \hspace{0.1cm}\text{Cal} \end{align}\]
Thus the required amount of heat is 200 cal + 1600 cal + 2000 cal + 10800 cal = 14600 cal .



9. Find the rms speed of Nitrogen at NTP. Density of N₂= 1.29 kg/m³ at NTP.

Solution :
density of nitrogen, \(\rho\) = 1.29 kg/m³
pressure , \(P\) = 1.01 \(\times\) 10\(^5\) Pa at NTP
rms speed of nitrogen, \(c_{rms}\) = ?

From the kinetic theory of gases, \[\begin{align} P&=\frac{1}{3}\rho \bar{c}^2\\ \bar{c}^2&=\frac{3P}{\rho}\\ &=\frac{3 \times 1.01 \times 10^{5}}{1.293}\\ c_{rms}=\sqrt{\bar{c}^2}&=\sqrt{\frac{3 \times 1.01 \times 10^{5}}{1.293}}\\ \therefore c_{rms}&=484.08 \hspace{0.1cm}\text{m/s} \end{align}\]
Hence the required solution is 484.08 m/s.




10. A mirror forms an erect image 30 cm from the object and twice its height. Where the mirror must be situated? What is the radius of curvature? Assuming the object be real, determine whether the mirror is convex or concave?

Solution :
magnification, m = -2 (\(\because\) the image is virtual)
Let mirror is situated at a distance x cm from the object, then

u = x
from the relation, \[m=\frac{v}{u}\] we find, v = -2x

Since the image is erect, it lies behind the mirror.
So the distance between the object and the image is 3x .
The given distance between the image and the object is 30 cm.
So, \[\begin{align} 3x&=30\\ x&=\frac{30}{3}\\ \therefore x&=10 \hspace{0.1cm} \text{cm} \end{align}\]
Then, \[v=-2x = -20 \hspace{0.1cm} \text{cm}\] Using lens formula, \[\begin{align} \frac{1}{f}&=\frac{1}{u}+\frac{1}{v}\\ &=\frac{1}{10}-\frac{1}{20}\\ &=\frac{1}{20}\\ \therefore f& = 20 \hspace{0.1cm} \text{cm} \end{align}\]
The radius of curvature is then \(R=2f=2 \times 20 = 40\) cm.
Since, f is positive, the mirror is concave.


11. An achromatic converging lens of mean focal length 40 cm is made by combining two lenses of different materials. If the dispersive power of the two lenses is in the ratio 1:3, find the focal length of each lens.

Solution :
mean focal length, F = 40 cm
Let f and f' be the focal legth of two lenses having \(\omega\) and \(\omega'\) as their dispersive powers.
\(\omega\) : \(\omega'\) = 1:3
The condition for achromatism is, \[\frac{\omega}{f}=-\frac{\omega'}{f'}\] Using this,
\[\begin{align} \frac{f}{f'}&=-\frac{\omega}{\omega'}\\ &=-\frac{1}{3}\\ f&=-\frac{f'}{3} \end{align}\]
Also, \[\begin{align} \frac{1}{F}&=\frac{1}{f}+\frac{1}{f'}\\ \frac{1}{40}&=\frac{1}{-\frac{f'}{3}}+\frac{1}{f'}\\ \frac{1}{40}&=-\frac{3}{f'}+\frac{1}{f'}\\ \frac{1}{40}&= -\frac{2}{f'}\\ f'&=-80 \hspace{0.1cm} \text{cm} \end{align}\]
And \[f=-\frac{f'}{3}=26.67 \hspace{0.1cm} \text{cm}\]
Thus the required focal lengths are - 80 cm and 26.67 cm.





12. The luminous efficiency of a lamp is 5 lumen/watt and its power is 88 watt. Calculate the illumination of a surface placed at a distance of 10 m, when light is falling normally.

Solution :
luminous efficiency = 5 lumen/watt
power , P = 88 watt
Illumination , I = ?
distance, r = 10 m

\[\text{luminous intensity(L)}=\frac{\text{luminous flux}}{\text{solid angle}}\] \[\text{luminous efficiency}=\frac{\text{luminous flux}}{\text{power}}\] \[L=Ir^2\]
From this relations, \[\begin{align} \text{luminous efficiency}&=\frac{\text{luminous flux}}{\text{power}}\\ 5 &=\frac{\text{L}\times 4\pi}{88}\\ 5& =\frac{I r^2 \times 4\pi}{88}\\ 5&=\frac{I \times 100 \times 4\pi}{88}\\ \frac{5\times 88}{400 \pi}&=I\\ \therefore I &=0.35 \hspace{0.1cm}lumen/m^2 \end{align}\] Thus the required solution is 0.35 lumen/m²

13. What is the equivalent of the combination of capacitors as shown in the figure? If C₁ = C₃=C₄=C₅ = 4μF and C₂ =10 µ F. Find the effective capacitance between A and B.


Solution :
In this circuit, when the battery is applied between the points A and B , b and c will be at the same potential and thus no charge flows through C₂.
Here, C₁ and C₅ are in series. Similarly, C₃ and C₄ are also in series. Let C be the effective capacitance of C₁ and C₅ and C' be the effective capacitance of C₃ and C₄.

\[\begin{align} C&=\frac{C_1 \times C_5}{C_1+C_5}\\ &=\frac{4\times 4}{4+4}\\ \therefore &=2 \hspace{0.1cm}\mu F \end{align}\]
\[\begin{align} C'&=\frac{C_3 \times C_4}{C_3+C_4}\\ &=\frac{4\times 4}{4+4}\\ \therefore &=2 \hspace{0.1cm}\mu F \end{align}\]
Since C and C' are in parallel, the effective capacitance becomes, (2 + 2) \(\mu\) F = 4 \(\mu\) F.

14. A hollow spherical conductor of radius 15 cm is charged to 6x10^-6C. Find the electric field strength at the surface of sphere, inside the sphere at 10 cm and at distance 20 cm from the sphere.

Solution :
radius of sphere, R = 15 cm = 0.15 m
charge, Q = 6x10^-6C
electric field strength at the surface, \(E_s\) = ?
electric field strenth inside the surface, \(E_i\) = ?
electric field outside the surface , \(E_o\) = ? at r = 20 cm = 0.20 m

Now, \[\begin{align} E_s&=\frac{Q}{4\pi \epsilon_0 R^2}\\ &=9 \times 10^9 \times \frac{6\times 10^{-6}} {(0.15)^2}\\ \therefore E_s &=2400000 \hspace{0.1cm} V/m \end{align}\]
There is no charge inside the hollow spherical conductor and thus \(E_i\) = 0.

And, \[\begin{align} E_o&=\frac{Q}{4\pi \epsilon_0 r^2}\\ &=9 \times 10^9 \times \frac{6\times 10^{-6}} {(0.20)^2}\\ \therefore E_o &= 1350000\hspace{0.1cm} V/m \end{align}\]
Thus the required solutions are 2.4 \(\times\) 10 ⁶ V/m, 0 and 1.35 \(\times\) 10⁶ V/m respectively.

15. A metallic strain gauge with K=2 is bonded to a steel member which is subjected to a stress of 10.5 × 10⁹ N/m². If modulus of elasticity for steel is 21 × 10¹² N/m², calculate the fractional change in the resistance of the gauge due to the applied stress. .

Solution :
K=2
stress = 10.5 × 10⁹ N/m²
modulus of elasticity, E = 21 × 10¹² N/m²
fractional change in resistance of the gauge, \(\frac{\Delta R}{R}\) = ?
Now,
\[\begin{align*} E&=\frac{\text{stress}}{\sigma}\\ \sigma&=\frac{\text{stress}}{E}\\ &=\frac{10.5 \times 10^9}{21\times 10^{12}}\\ \therefore \sigma&=5 \times 10^{-4} \hspace{0.1cm} N/m^2\\ \end{align*}\] Again,
\[\begin{align*} K&=\frac{\frac{\Delta R}{R}}{\sigma}\\ \frac{\Delta R}{R}&=\sigma K\\ &=5\times 10^{-4}\times 2\\ \therefore \frac{\Delta R}{R}&=10^{-3}\\ \end{align*}\]
Thus the fractional change in resistance is 10^-3.