Physics Test-21

Since acceleration \((a)=\frac{d v}{d t}\)

\(\mathrm{dv}=\mathrm{adt}\)

Integrate on both sides, we get

\(\int_u^v d v=\int_0^t a d t\)

So change in velocity \(=\) difference in velocity \(=v-u=\) at

Thus a - \(\mathrm{t}\) graph gives the difference in velocities.

Two point sources are regarded as just resolved when the principal diffraction maximum of one image coincides with the first minimum of the other. If one considers diffraction through a circular aperture, this translates into the expression

\(\theta=1.22 \frac{\lambda}{ d }=1.22 \times \frac{5000 \times 10^{-8}}{10}=1.22 \times 5 \times 10^{-6}=6 \times 10^{-6} rad\)

So, the closest is \(10^{-6} rad\).

Let the Mass of spacecraft = Mass of (P + Q + R)

From Newton's third law of motion, when the exhaust system produces a force of 1430 N, an equal and opposite force is exerted back on the spacecraft. Let this force be F.

∴F = 1430 N

As we know,

F =m × a

∴1430= Mass of spacecraft × a

⇒ 1430 = (1200 + 3150 + 2800) × a

⇒ 1430 = 7150× a

∴a = 0.2 m/s²

So, the acceleration of the spacecraft is 0.2 m/s².

When heat is given to particles, they gain energy. On gaining enough energy, particles start moving randomly to acquire kinetic energy and get mixed up with the particles present in air. It is a surface phenomenon.

We know that mercury does not evaporate at room temperature. It has its various applications like it is used in thermometers to check body temperatures also. We will see that mercury has a higher boiling point that is why it does not show the evaporation process at room temperature.

Work done is defined as \(\mathrm{W}=\mathrm{F.s}\)

where \(\mathrm{F}\) is the force applied and s is the displacement in the direction of force.

Hence initially work done \(\mathrm{W}_{1}=\mathrm{F.s}\)

Now, When \(\mathrm{F}_{2}=2 \mathrm{~F}\) and \(\mathrm{s}_{2}=2 \mathrm{~s}\)

Then, work done \(\mathrm{W}_{2}=\mathrm{F}_{2}. \mathrm{~s}_{2}\)

\(\Rightarrow \mathrm{W}_{2}=(2 \mathrm{~F}).(2 \mathrm{~s})\)

\(\Rightarrow \mathrm{W}_{2}=4 \mathrm{~W}_{1}\)

So, the work should be \(4\) times.

The resolving power of a telescope depends upon theDiameter of an objective.

R.P. of telescope\((\Delta \theta) \approx \frac{0.61 \lambda}{a}\)

Thus \(\Delta \theta\) will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if \(a\) is large. It is for this reason that for better resolution, a telescope must have a large diameter objective.

Given:

\(L=40\) cm \(=0.4\) m

\(W =10\) cm \(=0.1\) m

\(N =10\) turns

\(I =16\) A

\(\theta=60^{\circ}\)

\(B =0.60\) T

Where, \(N=\) number of turns in the coil, I = current in the loop, \(A=\) area enclosed by the loop, \(L=\) length of the loop, \(W=\) width of the loop, \(B=\) magnetic field intensity, and \(\theta=\) angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

The area enclosed by the loop is given as,

\(A = L \times W\)

\(\Rightarrow A =0.4 \times 0.1\)

\(\Rightarrow A =0.04\) m\(^{2}\)

So, the torque on the current-carrying rectangular loop is given as,

\(\tau =IAB \sin \theta\)

\(\Rightarrow \tau =16 \times 0.04 \times 0.6 \times \sin 60\)

\(\Rightarrow \tau=\frac{16 \times 4 \times 6 \times \sqrt{3}}{10 \times 10 \times 2}\)

\(\Rightarrow \tau=1.92 \sqrt{3}\) N-m

When the block is moving on a Rough surface Kinetic Frictional forces try to resist the movement.

Here, a block is moving with constant velocity, which means that no resultant force is acting on the block.

Thus, the frictional force is equal and opposite to the applied force and therefore 40 N force is acting on the block but in opposite direction to the external force.

Friction force = external force = 40 N

Thermal diffusivity is a Physical property of a substance.

In heat transfer analysis, thermal diffusivity is the thermal conductivity divided by density and specific heat capacity at constant pressure. It measures the rate of transfer of heat of a material from the hot end to the cold end. It has the SI derived unit m²/s.

Option (A) (Guanidine) is the strongest base because its conjugate acid is highly stable due to delocalization. It has 3 equivalent resonating structures.