A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
The bob of the simple pendulum will experience the centripetal acceleration provided by the circular motion of the car and the acceleration due to gravity.
Acceleration due to gravity = g
Centripetal acceleration = v2 / R
Where,
v is the uniform speed of the car
R is the radius of the track
Effective acceleration (g’) is given as
g‘ = vg2 + (v4 / R2)
Hence,
Time period, t = 2pvl / g’
= 2p v(l / vg2 + v4 / R2)
Therefore, its time period will be 2p v(l / vg2 + v4 / R2)
The marked price of an article is Rs 80 and it is sold at Rs 76, then the discount rate is:
(a) 5% (b) 95% (c) 10% (d) appx. 11%
(a) 5%
Explanation: Marked price = Rs. 80
Sold price = Rs.76
We know that,
Selling price = Marked price – Discount
Discount = Marked price – Selling price
Discount = Rs.80-Rs.76 = Rs.4
Discount % = 4/80 x 100 = 5%
Shyama purchases a scooter costing Rs 36,450 and the rate of sales tax is 9%, then the total amount paid by her is:
(a) Rs 36,490.50 (b) Rs 39,730.50 (c) Rs 36,454.50 (d) Rs 33,169.50
(b) Rs 39,730.50
Explanation: Scooter cost Rs.36450 at the rate of sales tax = 9%.
Total cost of scooter paid by Shyama = 9% of 36450 + 36450
= (9/100 × 36450) + 36450
= 3280.5 + 36450
= 39730.5
For calculation of interest compounded half yearly, keeping the principal same, which one of the following is true.
(a) Double the given annual rate and half the given number of years.
(b) Double the given annual rate as well as the given number of years.
(c) Half the given annual rate as well as the given number of years.
(d) Half the given annual rate and double the given number of years.
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Ashima took a loan of Rs 1,00,000 at 12% p.a. compounded half-yearly. She paid Rs.1,12,360. If (1.06)2 is equal to 1.1236, then the period for which she took the loan is:
(a) 2 years (b) 1 year (c) 6 months (d) 1(1/2) years
(b) 1 year
Explanation: P = Rs.100000, R = 12% per annum compounded half-yearly.
Amount = Rs.112360
Since we know,
A = P (1+R/100)T
112360 = 100000(1+12/100)T
112360/100000 = (1+12/100)T
(1.1236)1 = (1.12) T
If we compare the base terms, 1.1236 is approximately equal to 1.12
Hence, T = 1 year.
If a % is the discount per cent on a marked price x, then discount is
(a) (x/a) × 100 (b) (a/x) × 100 (c) x × (a/100) (d) 100/(x × a)
(c) x × (a/100)
(Discount = Discount% on Marked Price)
To gain 25% after allowing a discount of 10%, the shopkeeper must mark the price of the article which costs him Rs 360 as
(a) Rs 500 (b) Rs 450 (c) Rs 460 (d) Rs 486
(a) Rs 500
Explanation: Say, marked price = x
Cost price = Rs.360
As per the question;
x – [x×(10/100)] – [(25×360)/100] = 360
x – x/10 – 90 = 360
9x/10 = 360 + 90
9x = 4500
x = 500
If 90% of x is 315 km, then the value of x is
(a) 325 km (b) 350 km (c) 350 m (d) 325 m
(b) 350 km
Explanation: 90% of x is 315 km
90/100 × x = 315
X = 315 × 100/90 = 315 × 10/9 = 350
If marked price of an article is Rs 1,200 and the discount is 12% then the selling price of the article is (a) Rs 1,056 (b) Rs 1,344 (c) Rs 1,212 (d) Rs 1,188
(a) Rs 1,056
Explanation: Marked price = Rs.1200
Discount = 12%
Since, Discount = Discount% on Marked price
Discount price = 12% of 1200 = 12/100 × 1200 = 12 × 12 = 144
Selling price = Marked price-discount price = 1200 – 144 = Rs. 1056
The compound interest on Rs 50,000 at 4% per annum for 2 years compounded annually is (a) Rs 4,000 (b) Rs 4,080 (c) Rs 4,280 (d) Rs 4,050
(b) Rs 4,080
Explanation: P = Rs.50000, R = 4%, T = 2 years
A = P(1+R/100)T = 50000(1+4/100)2 = 50000(1+1/25)2
A = 50000(26/25)2 = 54080
Compound interest = A – P = 54080 – 50000 = Rs. 4080