Physics Test - 31

Using Lami's Theorem \(\therefore P: Q: R=\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}\)

\(\vec{R}=\vec{P}+\vec{Q}\)...(i)

\(\vec{S}=-\vec{P}+\vec{Q}\)....(ii)

Adding (i) and (ii) we get\(\vec{R}+\vec{S}=2 \vec{Q}\)....(iii)

Subtracting (i) and (ii) we get\(\vec{R}-\vec{S}=2 \vec{P}\)....(iv)

Divide eq. (iii) and (iv)

\(\frac{2 \vec{P}}{2 \vec{Q}}=\frac{\vec{R}-\vec{S}}{\vec{R}+\vec{S}}=\frac{\vec{R} \cdot \vec{R}-(\vec{S} \cdot \vec{R})}{\vec{R} \cdot \vec{R}+(\vec{S} \cdot \vec{R})}=1\)

\(\Rightarrow \frac{\vec{P}}{\vec{Q}}=1\)

Let the resistance of the wire be \(R\):

So power consumed \(p=\frac{v^{2}}{R}\)....(1)

When resistance \(R\) is divided in to \(4\) parts, resistance of each part is \(\frac{R}{4},\) so resulting resistance

\(\frac{1}{R^{\prime}}=\frac{4}{R}+\frac{4}{R}+\frac{4}{R}+\frac{4}{R}\)

Or \({R}={R} / 16\)

So now power \(=\frac{v^{2}}{R / 16}=16 p\)

So new power is \(16 p\) or \(3^{r d}\) option is the answer

\(\frac{{h}}{{h'}}=\frac{{n}_{2}}{{n}_{1}}\), where \(h\) is the real depth and \(h'\) is the apparent depth.

\(h=\frac{n_{2}}{n_{1}} \times h^{\prime}\)

Differentiating w.r.t. time

Amount of water drained \(=\) Area \(\times \frac{{dh'}}{{dt}}=\pi {R}^{2} {x}\)

Therefore,\(\frac{{dh}}{{dt}}=\frac{{n}_{2}}{{n}_{1}}\left(\pi {R}^{2}{x}\right)\)

Here assertion given is correct. The reason given for the assertion is correct but it is not the reason for the assertion.

Both A and B are true. In an iron-copper thermocouple, current flows from iron to copper since iron comes first in see back list.

Since the resistances \(4 \Omega,6 \Omega\) and \(5 \Omega\) are in parallel, therefore the voltage across them will be the same.

We know, \(P=\frac{V^{2}}{R}\)

\(P \times R=V^{2}\)

Across the \(5 \Omega\) resistance,

\(20 \times 5=100={V}^{2}\)

\({V}=10\) Volts

Current passing through the upper arm (whose equivalent resistance is \(10 \Omega)=\frac{10}{10}=1 A\) (Using \(V=IR\))

So, power dissipated by \(4 \Omega\) resistor \(=I^{2} R=4{W}\)

\(R^{2}=P^{2}+Q^{2}+2 P Q \cos \theta\).......(1)

\(4 R^{2}=P^{2}+4 Q^{2}+4 P Q \cos \theta\)....(2)

\(4 R^{2}=P^{2}+Q^{2}-2 P Q \cos \theta\)...(3)

On (1)+(2),\(5 R^{2}=2 P^{2}+2 Q^{2}\)....(4)

On (3) \(\times +\) (2),\(12 R^{2}=3 P^{2}+6 Q^{2}\)....(5)

\(2 P^{2}+2 Q^{2}-5 R^{2}\)....(6)

\(3 P^{2}+6 Q^{2}-12 R^{2}\)....(7)

By cross multiplication

\(\frac{P^{2}}{-24+30}=\frac{Q^{2}}{24-15}=\frac{R^{2}}{12-6}\)

\(\mathrm{mg}=\mathrm{T}_{2} \cos 60^{\circ}+\mathrm{T}_{1} \cos 30^{\circ}\).......(i)

\(\mathrm{T}_{1} \sin 30^{\circ}=\mathrm{T}_{2} \sin 60^{\circ}\)....(ii)

\(\Rightarrow {T}_{1} /{T}_{2}=\sqrt{3}\)....(iii)

\(\mathrm{mg}=\mathrm{T}_{2} \cos 60^{\circ}+\mathrm{T}_{1} \cos 30^{\circ}\)

\(\frac{240}{T_{2}}=\frac{1}{2}+\frac{T_{1}}{T_{2}} \times \frac{\sqrt{3}}{2}\)

Using eq (iii)

We get,\({T}_{2}=120 {kg}\) and \({T}_{1}=120 \sqrt{3}{kg}\)

Projectile is the name given to a body thrown with some initial velocity with the horizontal direction, and then allowed to move in two dimensions under the action of gravity alone, without being propelled by any engine or fuel. After projection, the object will move under the combined effect of two independent perpendicular velocities.

(i) Horizontal velocity, which remains constant throughout the motion (neglecting the air friction).

(ii) Vertical velocity which increases due to gravity. Initial value of this velocity at origin is zero.