Let where Then equals :
Use this Double Sin Formula
Question
Asked in: BITSAT-2018
From a sphere of mass M and radius R, a smaller sphere of radius is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :
Question
Asked in: BITSAT-2018
From a sphere of mass M and radius R, a smaller sphere of radius is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C
and P are
respectively, then the position vector of the orthocentre of this triangle, is :
If (10)^{9} + 2(11)^{1} (10)^{8} + 3(11)^{2} (10)^{7} +...... +10 (11)^{9} = k (10)^{9}, then k is equal to :
100
110
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :
8 A
10 A
12 A
14 A
In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively :
2.5 and 3
3.5 and 2
4.5 and 1
3 and 2.5
A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1. The battery terminals are connected to an external resistance ‘R’. The minimum value of ‘R’, so that a current passes through the battery to charge it is :
zero
In the circuit shown, the current in the 1 resistor is :
1.3 A ,from P to Q
0 A
0.13 A ,from Q to P
0.13 A ,from P to Q
answer is option 3)0.13A from Q to P
In the electric network shown, when no current flows through the resistor in the arm EB, the potential difference between the points A and D will be :
3V
4V
5V
6V
The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
10.04 Volt
Zero Volt
2.9 Volt
13.3 Volt
If and be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :
Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM. This Vanadium chloride has the formula :
(at. no. of V=23)
VCl_{2}
VCl_{4}
VCl_{3}
VCl_{5}
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is
6%
zero
1%
3%
This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statemens.
Statement - I : Higher the range, greater is the resistance of ammeter.
Statement - II : To increase the range of ammeter,additional shunt needs to be used across it.
Statement - I is false, Statement - II is true.
Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.
Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.
Statement - I is true, Statement - II is false.
A 10 V battery with internal resistance and a 15V battery with internal resistance are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :
11.9 V
12.5 V
13.1 V
24.5 V
A 3 olt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I i n the circuit will be
Two electric bulbs marked 25 W-220 V and 100 W - 220 V are connected in series to a 440 V supply .Which of the bulbs will fuse?
Three capacitances, each of 3 F, are provided. These cannot be combined to provide the resultant capacitance of :
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about
150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be:
+670 kC
- 670 kC
- 680 kC
+ 680 kC
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :