GATE Questions

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When the switch is ON, in a tube light choke is nothing but the coil/ballast ( inductor) which is used to induce the high voltage across it. as we know that inductor has property to induce high voltage for the brief period of time, this high voltage is required to ionize the gases in the starter.

Then gas inside the starter gets ionized due to this full voltage and heats the bimetallic strip that is caused to be bent to connect to the fixed contact. Current starts flowing through the starter. Although the ionization potential of the neon is the little bit more than that of the argon but still due to small electrode gap high voltage gradient is appeared in the neon bulb and hence glow discharge is started first in a starter.

As voltage gets reduced due to the current causes a voltage drop across the inductor, the strip cools and breaks away from the fixed contact. At that moment a large L di/dt voltage surge comes across the inductor at the time of breaking.

This high valued surge comes across the tube light electrodes and strikes penning mixture (mixture argon gas and mercury vapor).

Gas discharge process continues and current gets the path to flow through the tube light gas only due to low resistance as compared to the resistance of starter.

The relationship is xy : x+1y+1

Now , 64 = 43

So, required number = 4+13+1 = 54 = 625

$$\begin{gathered} 23x35x34x?=3174 \\ 3174x10/3 \\ = 10580 \end{gathered}$$

I      II     III

65   58   100

65 -------- 7
100 ------ ?  => 10.769%

Let the time taken by train be 't' hrs.

Then,

40t = 35t + 1560

40t = 35t + 35/4

t = 7/4 hrs

Therefore, Required length of the total journey d = s x t

= 40 x 7/4

= 70 kms.

Let the number be x. Then,

3x2 - 4x = x + 50

3x2- 5x - 50 = 0

(3x + 10)(x - 5) = 0

x = 5

By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.

Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 5780
= x= 5780/2.89 = 2000

Therefore, amount invested by Raghu is Rs. 2000.