Question

Four point masses, each of 1 kg are joined together by string, which form
diagonal 0.707 m. The square is placed on a rotating table, which is rotated at
5 rps. The tension in the string is:
> 246.8 N
a) 2.468 N
b) 0.2468 N
d) 2468 N​

Answer 1

Explanation:

hope you get the solution

Answer 2

Given that,

Each mass = 1 kg

Diagonal distance = 0.707

Angular frequency = 5 rps

Lets take length of square is a meter diagonal will be $\sqrt{2}a$

We need to calculate the value of a

Using formula of diagonal distance

$\sqrt{2}a =d$

Put the value into the formula

$\sqrt{2}a=0.707$

$a=\dfrac{0.707}{\sqrt{2}}$

$a=0.5\ m$

We need to calculate the angular frequency in rad/s

Using formula angular frequency

$\omega=2\pi\times5$

$\omega=10\pi\ rad/s$

We know that,

Radius of revolution is equal to the half of diagonal

$r=\dfrac{\sqrt{2}a}{2}$

$r=\dfrac{a}{\sqrt{2}}$

We need to calculate the tension in the string

Using balance equation

$2T\cos\theta=mr\omega^2$

Put the value in to the formula

$2T\cos45=1\times\dfrac{0.5}{\sqrt{2}}\times(10\pi)^2$

$\dfrac{2T}{\sqrt{2}}=1\times100\pi^2\times\dfrac{0.5}{\sqrt{2}}$

$T=\dfrac{1\times100\pi^2\times\dfrac{0.5}{\sqrt{2}}}{\sqrt{2}}$

$T=246.8\ N$

Hence, The tension in the string is 246.8 N.

(c) is correct option.

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Topic : tension

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