A stone of mass m tied to the end of a string is revolving in a vertical circle of radius R. The net force at the lowest and highest points of the circle directed vertically downwards are: (choose the correct alternative).
T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
The net force at the lowest point is (mg – T1) and the net force at the highest point is (mg + T2). Therefore option (a) is correct.
Since mg and T1 are in mutually opposite directions at the lowest point and mg and T2 are in the same direction at the highest point.
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawnmower than to push it,
(d) a cricketer moves his hands backwards while holding a catch
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Mass of the stone = 0.25 kg
Radius, r = 1.5 m
Number of revolution in a second, n= 40/60 = (?) rev/sec
The angular velocity, ? = 2pn = 2 x 3.14 x (?)
The tension on the string provides the necessary centripetal force
T = m?2r
T = 0.25 x 1.5 x [2 x 3.14 x (?)]2
= 6.57 N
Maximum tension on the string, Tmax= 200 N
v2max = (Tmax x r)/m
= ( 200 x 1.5)/0.25 = 1200
vmax = 34. 6 m/s
Therefore, the maximum speed of the stone is 34.64 m/s
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
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)
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%
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
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.
Hey are you a science teacher
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