Physics Test - 10

As,

\(h=u t+\frac{1}{2} g t^{2}\)

Here, \(u=0, g=10 m s^{-2}, t=4 s\)

\(\therefore h=0 \times 4+\frac{1}{2} \times 10 \times 4^{2}=80\)m

The distance \(s_1\) covered by the car during the time it is accelerated is given by \(2 a s_1=v^2\), which gives \(s_1=\frac{v^2}{2 a}\). The distance \(s_2\) covered during the time the car is decelerated is similarly given by \(s_2=\frac{\mathrm{v}^2}{2 \beta}\)

Therefore, the total distance covered is

\(s=s_1+s_2=\frac{v^2}{2}\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots \text { (i) }\)

If \(t_1\) is the time of acceleration and \(t_2\) that of deceleration, then \(v=a t_1=\beta t_2 \Rightarrow \mathrm{t}_1=\frac{\mathrm{v}}{a}\) and \(\mathrm{t}_2=\frac{\mathrm{v}}{\beta}\)

Therefore, the total time taken is

\(t=t_1+t_2=v\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots\)(ii)

From (i) and (ii), the average speed of the car is given by,

\(\frac{\text { Total distance }}{\text { Total time }} \frac{s}{t}=\frac{v}{2}\)

Root mean square velocity of Gaseous particles: Root mean square velocity (RMS value) is the square root of the mean of squares of the velocity of individual gas molecules. It is given by,

\(V_{r m s}=\sqrt{\frac{3 R T}{M}}\)

Average velocity: It is the arithmetic mean of the velocities of different molecules of a gas at a given temperature. It is given by,

\(V_{a v}=\sqrt{\frac{8 R T}{\pi M}}\)

Most probable velocity: It is the velocity possessed by maximum fraction of molecules at the same temperature. It is given by,

\(V_{p}=\sqrt{\frac{2 R T}{M}}\)

The required ratio between these parameters are,

\(\sqrt{\frac{3 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{2 R T}{M}}\)

By Canceling all constant terms from the above expression we have,

\(\sqrt{3}: \sqrt{\frac{8}{\pi}} : \sqrt{2}\)

Or

1.732:1.596:1.414

A positive charge attracts a negative charge and repels the positive charge. A negative charge attracts a positive charge and repels a negative charge. It means Like charges repel each other and unlike charges attract each other. All the charges do not attract all the other charges always.

If a transformer of an audio amplifier has an input impedance of 8Ω output impedance of 8 kΩ, the primary and secondary turns of this transformer connected between the output of the amplifier and to loudspeaker should have the ratio will be 1:32.

Given:

Electrostatic force of repulsion, \(\mathrm{F}=3.7 \times 10^{-9} \mathrm{~N}\)

Let charge is \(q_{1}=q_{2}=q\)

Distance between two charges, \(r=5 \mathring{A}=5 \times 10^{-10} \mathrm{~m}\)

We have to find, the number of electrons(n) missing.

Using Coulomb's law,

\(\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \)

\(\Rightarrow 3.7 \times 10^{-9}=9 \times 10^{9} \times \frac{\mathrm{q} \times \mathrm{q}}{\left(5 \times 10^{-10}\right)^{2}} \)

\(\Rightarrow \mathrm{q}^{2}=\frac{3.7 \times 10^{-9} \times 25 \times 10^{-20}}{9 \times 10^{9}} \)

\(\Rightarrow \mathrm{q}^{2}=10.28 \times 10^{-38} \)

\(\Rightarrow q=3.2 \times 10^{-19} \text { Coulomb }\)

As, we know that:

\(q=\text { ne } \)

\(\therefore n=\frac{q}{e} \)

\(=\frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} \)

\(=2\)

Therefore, the number of electrons missing is \(n=2\).

In a transverse waves particle of the medium vibrate up and down in the vertical direction whereas it is propagating along the horizontal direction. So, in a transverse wave, a crest is a part where particle rises from its mean position and has the maximum positive displacement whereas trough is a part where particle dips below mean position and has the maximum negative displacement.

Archimedes principle:

• It states that a body when wholly or partially immersed in liquid experiences an upward thrust which is equal to the volume of the liquid.
• Archimedes Principle is also known as thephysical law of buoyancy.
• Purity of gold in the king's crown,Archimedes determine after discovering the Archimedes principle.
• Hydrometer, Ships, and Submarines work on the Archimedes Principle.
• A Ship/boat floats on the basis of theArchimedes Principle.

As the c.m. of three particles is at \(2,2,2\)

\(\therefore \text { The total mass }=1+2+3=6 \mathrm{~kg}\)

Now consider the \(4 \mathrm{~kg}\) mass at the position \((\mathrm{x}, \mathrm{y}, \mathrm{z})\)

Now centre of mass of total system at \(0,0,0\)

\(\therefore \frac{6 \times 2+4 x}{10}=0\)

\(\Rightarrow 12=-4 x\)

\(\Rightarrow x=-3\)

Similarly, \( \frac{6 \times 2+4 y}{12}=0\)

\(\Rightarrow y=-3\)

Similarly, \(\frac{6 \times 2+4 z}{12}=0\)

\(\Rightarrow z=-3\)

From the above we can conclude that

\(x=-3, y=-3, z=-3\)