A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration '$a$' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration '$a$' vertically. The tension in the string is equal to -
- $m \sqrt{g^{2} + a^{2}}$
- $m \sqrt{g^{2} + a^{2}} - ma$
- $m \sqrt{g^{2} + a^{2}} + ma$
- $m (g + a)$
The Correct Option is C
Solution and Explanation
$T - m \sqrt{ a ^{2}+ g ^{2}}= ma$
$T = ma + m \sqrt{ a ^{2}+ g ^{2}}$
Top Questions on tension
- Two monkeys off mass 10 kg and 8 kg are moving along a vertical light rope the former climbing up with an acceleration of 2 m/second square while the latter coming down with a uniform velocity of 2 m/sec square find the tension in the rope at the fixed support
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A simple pendulum is made of a metal wire of length \( L \), area of cross-section \( A \), material of Young's modulus \( Y \), and a bob of mass \( m \). This pendulum is hung in a bus moving with a uniform speed \( V \) on a horizontal circular road of radius \( R \). The elongation in the wire is:
The velocities of air above and below the surfaces of a flying aeroplane wing are 50 m/s and 40 m/s respectively. If the area of the wing is 10 m² and the mass of the aeroplane is 500 kg, then as time passes by (density of air = 1.3 kg/m³), the aeroplane will:
Questions Asked in BITSAT exam
- Let \( ABC \) be a triangle and \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of \( A, B, C \) respectively. Let \( D \) divide \( BC \) in the ratio \( 3:1 \) internally and \( E \) divide \( AD \) in the ratio \( 4:1 \) internally. Let \( BE \) meet \( AC \) in \( F \). If \( E \) divides \( BF \) in the ratio \( 3:2 \) internally then the position vector of \( F \) is:
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- Let \( \mathbf{a} = \hat{i} - \hat{k}, \mathbf{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \mathbf{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \). Then, \( [\mathbf{a} \, \mathbf{b} \, \mathbf{c}] \) depends on:}
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- Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:
- The magnitude of projection of the line joining \( (3,4,5) \) and \( (4,6,3) \) on the line joining \( (-1,2,4) \) and \( (1,0,5) \) is:
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- The angle between the lines whose direction cosines are given by the equations \( 3l + m + 5n = 0 \) and \( 6m - 2n + 5l = 0 \) is:
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Concepts Used:
Tension
A force working along the length of a medium, especially if this force is carried by a flexible medium like cable or rope is called tension. The flexible cords which bear muscle forces to other parts of the body are called tendons.
Net force = 𝐹𝑛𝑒𝑡 = 𝑇−𝑊=0,
where,
T and W are the magnitudes of the tension and weight and their signs indicate a direction, be up-front positive here.
Read More: Tension Formula