Look at the picture of the tension protractor (in the information section). Supp

Question

Look at the picture of the tension protractor (in the information section).

Suppose mass (m = 460 g) hanging in the middle divides the string (length L = 43 cm) exactly in half (and the tension in the two strings are the same), and the distance between the ends of the string (on top) is 24.1 cm.

3. Tension Protractor a) The set-up appears as shown to the right. The two strings are attached to the tension protractor and the mass holder. Be sure the hanging mass does not touch the table b) The tension protractor has two scales: the inner (white) scale gives the tension in the string (in Newtons). You can read this value to the nearest tenth of a Newton. The outer (black) scale gives the angle between the rope and the horizontal (01 and 2). You can read this scale to the nearest degree. A close-up of the two scales appears to the right OISNa c) Before you begin - Disconnect the strings, and hang a weight from each scale separately. The outer scale should read 90°; if it not, loosen the set screw and turn the outer scale until it does 90 - Now remove the weight. The inner (Newton) scale should read 0; if it does not, loosen the set screw and reset the inner dial

Explanation / Answer

(b)
When it is shifted to Left,
T1*cos(1) = T2*cos(2)
1 > 2
cos(1) < cos(2)
Therefore,
T1 > T2
T1 Increases, T2 decreases.

Correct option - (3)

(c)
Initially when at Rest,
2*T*cos() = m*g
When it is accelerated Upwards,
2*T*cos() - m*g = m*a
2*T*cos() = m*(a + g)

So. Tension increases for both the string.

When It is Accelerated to the right, The mass will tend to come towards Left and the situation will be same as above in part(b),Therefore
T1 > T2
T1 Increases, T2 decreases.

Correct option - (4)

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