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Belt Rope and Chain Drives Questions and Answers

1.

Which of the following drives offers the highest power transmission efficiency?

Exam Year(s)

DSSSB - 2023,

Answer: A

The rope drive offers the highest power transmission efficiency in the list, as the key states. Rope drives are used to transmit large power over long distances with many ropes in grooved pulleys. The flat belt, circular belt and V-belt drives suffer from more slip or lower efficiency in comparison, so the key chooses the rope drive.

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Answer: B

The advantage of a chain drive is that it transmits more power than belts. A chain engages the sprocket teeth with no slip and it can carry heavy loads, and so it can transmit large powers at a constant velocity ratio. It is not claimed to be cheaper than a belt drive, it needs regular maintenance and lubrication, and it needs accurate mounting and alignment, so options A, C and D are wrong.

Because a chain is a positive drive it keeps a constant velocity ratio, unlike belts that slip or creep, and it can work over a longer centre distance than a gear pair.

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3.

\(180 - 2\sin^{-1}\left(\dfrac{r_1 - r_2}{x}\right)\)

The above expression belongs to

Exam Year(s)

DSSSB - 2023,

Answer: C

The expression \(180 - 2\sin^{-1}\frac{r_1 - r_2}{x}\) gives the angle of contact for an open belt drive. In an open belt drive the two pulleys of radii \(r_1\) and \(r_2\) are at a centre distance x, and the angle of contact on the smaller pulley is the angle subtended by the belt, which is \(180^{\circ} - 2\alpha\) where \(\sin\alpha = \frac{r_1 - r_2}{x}\). In a crossed belt drive the angle is \(180^{\circ} + 2\alpha\) with \(\sin\alpha = \frac{r_1 + r_2}{x}\), and the key gives the angle of contact rather than the angle of lap for the open belt.

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Answer: D

The statement that a fibre rope drive is suitable for indoor applications is not correct in the key. Fibre ropes are used for transmitting power over long distances, mainly in open places. A fibre rope drive has high mechanical efficiency, is smooth and quiet, and tolerates a little misalignment of the shafts, so the other three statements are correct.

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Answer: D

The approximate length of an open belt is \[ L = \pi(r_1 + r_2) + \frac{(r_1 - r_2)^{2}}{x} + 2x \] where \(r_1\) and \(r_2\) are the radii of the pulleys and x is the distance between their centres. The straight portions of the belt each add about x, giving 2x, and the wrapped portions on the two pulleys add \(\pi(r_1 + r_2)\) plus a correction term depending on the difference of the radii. The other options have the square in the wrong place, or the sum of the radii in the correction term.

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6.

A flat belt of negligible mass and thickness drives an output pulley of 0.6 m diameter, constantly rotating at 500 rpm without slip. The arc of contact is 190° for this pulley, while it is 170° for the smaller input pulley. The coefficient of friction between the belt and pulleys is 0.30. The measured slack-side tension is 250 N. Neglecting other losses, the power transmitted by the belt drive, in kW, is nearest to

(Take π = 3.14)

Exam Year(s)

GATE AG - 2026,

Answer: B

The power transmitted is about 5.6 kW.

A flat belt slips first on the pulley with the smaller arc of contact, so the tension ratio is limited by the 170° arc of the input pulley, not the 190° arc of the output pulley:

\[ \begin{aligned} \theta &= 170 \times \frac{3.14}{180} = 2.966\ \text{rad} \\ \frac{T_{1}}{T_{2}} &= e^{\mu\theta} = e^{0.30 \times 2.966} = 2.434 \\ T_{1} &= 2.434 \times 250 = 608.6\ \text{N} \end{aligned} \]

The belt speed follows from the output pulley, which turns at 500 rpm without slip:

\[ \begin{aligned} v &= \frac{\pi D N}{60} = \frac{3.14(0.6)(500)}{60} = 15.7\ \text{m/s} \\ P &= (T_{1} - T_{2})v = 358.6 \times 15.7 \\ &\approx 5630\ \text{W} \end{aligned} \]

So the power is nearest to 5.6 kW. Using the 190° arc of the larger pulley overstates the tension ratio and gives about 6.7 kW, the trap behind the larger options; 4.8 kW understates the tight-side tension.

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7.

The velocity ratio of belt drive can be expressed mathematically as :

Where :
N1 = Speed of driver in r.p.m
N2 = Speed of follower in r.p.m
d1 = Diameter of driver
d2 = Diameter of follower

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

The velocity ratio is \(\frac{N_2}{N_1} = \frac{d_1}{d_2}\).

If the belt does not slip, every point on it moves at the same linear speed, so the rim speeds of the two pulleys are equal:

\[ \begin{aligned} \pi d_1 N_1 &= \pi d_2 N_2 \\ \frac{N_2}{N_1} &= \frac{d_1}{d_2} \end{aligned} \]

The speeds are inversely proportional to the diameters, so the smaller follower runs faster. The \(\pi\) appears on both sides and cancels, which rules out the two options that keep it, and \(\frac{d_2}{d_1}\) inverts the ratio.

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8.

The creep in a belt drive is due to the :

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

Creep is due to the unequal tension on the tight and slack sides of the belt.

The tight side of a belt is stretched more than the slack side. As a part of the belt passes round the driving pulley from the tight side to the slack side, it contracts, and as it passes round the driven pulley it stretches again. This relative movement between belt and pulley is called creep, and it makes the driven pulley turn slightly slower than the diameters alone would give.

Creep is caused by the belt's elasticity under unequal tension, not by the material of the pulleys or their size.

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9.

The groove angle of the pulley for V-belt drive is usually:

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The groove angle is usually 32° - 38°.

A V-belt pulley groove is made with an included angle of about 30 to 40 degrees, commonly specified between 32 and 38 degrees. The belt wedges into the groove, which increases the normal reaction and so the friction it can transmit.

Angles of 20 to 32 degrees would make the belt jam in the groove, and 38 to 45 degrees would lose the wedging action that gives a V-belt its grip.

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10.

If T1 and T2 are the tensions on tight and slack sides of belt in Newtons and v is the velocity in m/sec, the power transmitted is given in Watt as

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

The power is \((T_{1} - T_{2})v\) watts.

The net force driving the pulley is the difference of the tight-side and slack-side tensions. Power is force times velocity, so with tensions in newtons and velocity in m/s the power comes out directly in watts.

\[ P = (T_{1} - T_{2})\,v \]

Dividing by 4500 or 75 converts to horsepower in old gravitational units (kgf m/min or kgf m/s), and dividing by 60 belongs to formulas with speed in revolutions per minute. None of these is needed for watts.

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