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3801.

What is dispersion ?

Answer»

Solution :The PHENOMENON of splitting up of a POLYCHROMATIC beam of LIGHT into it.s COLORS is called DISPERSION.
Discussion
3802.

Let two cell of e.m.f's E_1 and E_2 and internal resistances r_1 and r_2 be connected in parallel to a circuit with an external resistance R. Find the value of the current I through the given resistor .

Answer»

SOLUTION :ELECTRIC cells connected in parallel to an EXTERNAL resistor. The CURRENT through the given resistor

`I = ((E_1)/(r_1) + (E_2)/(r_2))/(1+ R[1/r_1 + 1/r_2]) = ((E_1 r_2 + E_2 r_1)/(r_1 r_2))/(1+ (R[r_1 +r_2])/(r_1 r_2))`
`I= (E_1 r_2 + E_2 r_1)/(r_1 r_2+ R[r_1 + r_2])`
Discussion
3803.

water rises to height of 6cm in a capillary tube of radius r. If the radius of the capillary is made 2r/3, then what is the height to which the liquid rises in the capillarity?

Answer»

SOLUTION :`h_2 = (h_1 r_1)/(3/4 R) = 3.3 XX 4/3 = 4.4 CM`
Discussion
3804.

Many types of race cars depend on negative lift (or down-force) to push them down against the track surface so they can take turns quickly without sliding out into the track wall . Part of the negative lift is the ground force , which is a force due to the airflow beneath the car . As the race car in Fig moves forward at 27.25 m/s , air is forced to flow over and under the car . The air forced to flow under the car . The air forced to flow under the car enters through a vertical cross-sectional area is A_(1) = 0.0310 m^(2) . Treat this flow as steady flow through a stationary horizontal pipe that decreases in cross-sectional area from A_(0) to A_(1) (Fig) (a) At the moment it passes through A_(0) , the air is at atmospheric pressure p_(0) . At what pressure p_(0) . At what pressure p_(1) is the air as it moves through A_(1) ?

Answer»

<P>

Solution :From Eq. , we can WRITE
`A_(0) v_(0) = A_(1) v_(1)` ,
or `v_(1) = v_(0) - (A_(0))/(A_(1)) "" (14-52)`
Substituting Eq.14-52 into Eq.14-51 and rearranginggive us
`p_(1) = p_(0) - (1)/(2) RHO v_(0)^(2) ((A_(0)^(2))/(A_(1)^(2)) - 1) "" (14-53)`
The speed of the air as it enters `A_(0)` at the front of the CAR is equal to 27.25 m/s , the speed of the car as it moves forward through the air . Substituting this speed , the air density `rho = 1.21 KG//m^(3)` , and the values for `A_(0)` and `A_(1)` into Eq. , we find
`p_(1) = p_(0)- (1)/(2) (1.21 kg //m^(3)) (27.25 m//s)^(2) [((0.0330 m^(2))^(2))/((0.0310 m^(2))^(2)) - 1]`
`p_(1) = p_(0) - 59.838 Pa = p_(0) - 59.8 Pa`
Discussion
3805.

A liquid rises in a capillary tube when the angle of contact is

Answer»

An acute ANGLE
An ABUSE angle
A RIGHT angle
`PIE` radian

Answer :A
Discussion
3806.

A sphere of mass m and radius r rolls without slipping over a tunnel of width d=(8r)/(5) as shown in the figure. (in the figure the sphere rolls perpendicular to the plane of the page) the velocity of centre of mass of sphere is v directed into the plane of the page. the maximum speed of a point on the sphere is

Answer»

`(4v)/(3)`
`(8v)/(3)`
`(6V)/(3)`
`(3v)/(2)`

Solution :
`v=omegaxx(3R)/(5)impliesomega=(5V)/(3r)`
`v_(B)=omegaxx(8r)/(5)=(5v)/(3r)XX(8r)/(5)=(8v)/(3)`
Discussion
3807.

A force couple is the term used for a system of two equal antiparallel forces, the arm of the couple is the shortest distance between the forces. Prove that the torque is equal to the product of the magnitude of the force and the arm no matter what is the position of the point with respect to which the torque is determined.

Answer»


Solution :From an arbitrary point o draw an axis perpendicular TOTHE plane in which the forces LIE (Fig). The moment of the force COUPLE

(the torque) is equal to the algebraic sum of the moments of each force about the axis. We have
`M=F_1a+F_2(a+d)=F_1a+F_2a+F_2d`
However, `F_1=F and F_2 =-F` , so M =-FD. The minus sign shows that this couple makes a left-hand screw about the axis.
Discussion
3808.

In a current carrying conductor, the ratio of the electric field and the current density at a point is called

Answer»

RESISTIVITY
conductivity
resistance
MOBILITY

SOLUTION :As PER vector form of Ohm.s law `vecE=rho vecJ`
Discussion
3809.

The ratio of maximum acceleration to the maximum velocity of a particle performing S.H.M. equals to the-

Answer»

Amplitude
ANGULAR velocity
Square of amplitude
Square of angular velocity

Answer :B
Discussion
3810.

X-rays of wavelength 22 pm are scattered from a carbon target at an angle of85^(@) to the incident beam The Compton shift forX-rays is(cos 85^(@) = 0.088)

Answer»

2.2 PM
1.1 pm
0.55 pm
4.4 pm

ANSWER :A
Discussion
3811.

A crystal diode having internal resistance r_f =20Omega is used for half-wave rectification. if the applied voltage v = 50 sin omegat and load resistance R_L = 800Omega find:a.c. power and d.c. power output

Answer»

SOLUTION :v=50 sin `OMEGAT`therefore` maximum voltage, `V_m=5V`a.c.power input`=(I_(RMS))^2XX(r_f+R_L)=(30.5/1000)xx(20+800)=0.763`watt d.c.power OUTPUT `I_(dc)^2xxR_L=(19.4/1000)^2xx800=0.301"watt"`
Discussion
3812.

A crystal diode having internal resistance r_f =20Omega is used for half-wave rectification. if the applied voltage v = 50 sin omegat and load resistance R_L = 800Omega find: d.c. output voltage

Answer»

SOLUTION :v=50 SIN `OMEGAT`therefore` MAXIMUM voltage, `V_m=5V` d.c. OUTPUT voltage `I_(dc)R_L=19.4mAxx800Omega=15.52"volts"`
Discussion
3813.

A crystal diode having internal resistance r_f =20Omega is used for half-wave rectification. if the applied voltage v = 50 sin omegat and load resistance R_L = 800Omega find: efficiency of rectification

Answer»

SOLUTION :v=50 sin `omegat`THEREFORE` MAXIMUM VOLTAGE, `V_m=5V` efficiency of RECTIFICATION `0.301/0.763xx100=39.5%`
Discussion
3814.

According to Maxwell’s hypothesis, a changing electric field gives rise to

Answer»

an e.m.f
electric CURRENT
MAGNETIC field
pressure gradient

Answer :C
Discussion
3815.

Keeping the velocity of projection constant, the angle of projection is increased from 0^(@) to 90^(@). Then the maximum height of the projectile

Answer»

GOES on increasing up to `90^(@)`
DECREASES up to `90^(@)`
increases up to `45^(@)` anddecreases beyond it
decrease up to `45^(@)` and increases beyond it

Answer :A
Discussion
3816.

A crystal diode having internal resistance r_f =20Omega is used for half-wave rectification. if the applied voltage v = 50 sin omegat and load resistance R_L = 800Omega find: I_m,I_(dc), I_(rms)

Answer»

SOLUTION :`v=50 SINOMEGAT therefore` maximum VOLTAGE, `V_m=5V I=V_m/(r_f+R_L)=50/(20+800)=0.061mA I_(dc)=I_m//pi=61//pi=19.4mAI_(rms)=I_m//2=61//2=30.5mA`
Discussion
3817.

What is order of conductivity of conductor?

Answer»


ANSWER :`10^(8)OMEGA(-1)m^(-1)`
Discussion
3818.

A : The film which appears bright in reflected system will appear dark in the transmitted light and vice-versa. R : In interference, the conditions for film to appear bright or dark in reflected light are just reverse to those in the transmitted light.

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Both A and R are true and R is the correct EXPLANATION of A
Both A and R are true and R is not the correct explanation of A
A is true and R is false
Both A and R are false

Answer :A
Discussion
3819.

Both radiowaves and gamma rays are transverse in nature and moving with same speed in free space. Then in what aspect are they different ?

Answer»

Solution :i. Radiowaves have ATOMIC origin while gamma RAYS NUCLEAR origin.
ii.Penetrating power of radiowaves is SMALL while that of gamma rays is large.
Discussion
3820.

In the Young.s double slit experiment , the intensities at two points P_(1)andP_(2) on the screen are respectively I_(1) and I_(2) .If P_(1) is located at the centre of a bright fringe and P_(2) is located at a distance equal to quarter of fringewidth from P_(1) . then I_(1)//I_(2) is

Answer»

2
`1//2`
4
16

Answer :A
Discussion
3821.

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 600 nm is incident on it . The energy band gap (in eV) for the semiconductor is

Answer»

a. `1.50`
B. `0.75`
C. `2.06`
d. `0.90`

ANSWER :C
Discussion
3822.

Explain briefly how bar magnets act as equivalent solenoids.

Answer»

Solution :
The magnitude of magnetic field at a point 'P' due TOA circular element of thickness 'dx' is given by `dB=((mu_(o))/(4pi))(2pindxIa^(2))/(((r-X)^(2)+a^(2))^(1//2))`
The magnitude of total magnetic field is given by
`B=((mu_(o))/(4pi))2pinIa^(2)int_(-l)^(l)(dx)/([(r-x)^(2)+a^(2)]^(3//2))`
For `r gt gt a ` and r `gt gtl, [(r-x)^(2)+a^(2)]^(3//2)=r^(3)`
`:. "" B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))int_(-l)^(l)dx "" int_(-l)^(l)dx=(x)_(-l)^(l)`
i.e., `"" B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))2L "" = l +l = 2l`
We note that the product n `(2l) I (pia^(2))` = Magnetic moment 'm'
`:. "" B = ((mu_(o))/(4pi)) (2m)/(r^(3)) "" --(1)`
This expression is similar to the magnetic field at a point on the axis of a short magnet. Therefore a bar magnet and a solenoid produce similar magnetic FIELDS.
Discussion
3823.

Two radiations with photon energies 0.9 eV and 3.3eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV , then the ration of maximum speeds of emitted electrons will be

Answer»

`1 : 4`
`1:3`
`1:1`
`1: 9`

SOLUTION : `v_(1_("MAX"))=SQRT((2(hv-phi))/(m))= sqrt((2(0.9-0.6))/(m)) =sqrt((2(0.3))/(m))`
`v_(2_"max")=sqrt(((hv-phi))/(m)) = sqrt((2(3.3-0.6))/(m)) = sqrt((2(2.7))/(m))`
`v_(1_(max)) = sqrt((2(0.3))/(m)) xx sqrt((m)/(2.7)) = sqrt(0.3 / 2.7 )= 1/3`
`(v_1 : v_2 ) _(max) = 1:3`
Discussion
3824.

The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersive power of the material of the lens.

Answer»


Solution :The focal LENGTH of lens is given by,
`(1)/(F) = (mu - 1). (1)/(R_1) - (1)/(R_2)`
`rArr (mu - 1) = (1)/(f)xx (1)/((1)/(R_1) - (1)/(R_2))`
`=(k)/(f)….. (i)`
so, `mu_r-1 = (k)/(100) … (ii)`
`mu_y-1 = (k)/(98)…. (ii)`
and `mu_(upsilon) -1 = (k)/(96) ..... (ii)`
So, Dispersive POWER ` =OMEGA`
`=(mu_(upsilon)j - mu_r)/(mu_y -1)`
`=(mu_(upsilon) -1) - (mu_r -1))/((mu_y-1))`
` =((k)/(96) - (k)/(100))/((k)/(98)) = (98xx 4)/(9600)`
=0.0408
Discussion
3825.

Wave theory cannot explain the phenomena of A. Polarization, B.Diffraction C. Compton effect , D. Photoelectric effect Which of the following is correct ?

Answer»

A and B
B and D
C and D
D and A

Solution :( c) :Wave theory explain the phenomena of interference, DIFFRACTION and polarization but it can't explain the photoelectric EFFECT and COMPTON effect.
Discussion
3826.

What must be the momentum of a particle with mass m so that the total energy of the particle is 4.00 times its rest energy?

Answer»


ANSWER :3.87 MC
Discussion
3827.

Speed of electromagnetic waves v in a medium of permittivity in and magnetic permeability mu is given as per relation.

Answer»

`V=SQRT(muin)`
`v=sqrt((in)/(MU))`
`v=sqrt((mu)/(in))`
`v=(1)/(sqrt(mu in))`

ANSWER :D
Discussion
3828.

The value of mu_(0) Is equal to ............ .

Answer»


ANSWER :`4 pi XX 10^(7) `Hm
Discussion
3829.

A 1000 kg engine pulls a train of 4 wagons each of 2500 kg along a horizontal railway track. If the engine exerts a force of 50000 N on wagons and track offers force of friction 10000 N, then calculate (i) The net accelerating force (ii) The acceleration of the train (iii) The force of wagon-1 on wagon-2

Answer»

Solution :
FORCE EXERTED by the engine is `F_(1)=50,000 N `
Frictional force exerted by the track, `F_(2)= -10000 N`
Negative sign shows that frictionis opposite to the directionof motion.
(i) Therefore, net accelerating force `(VECF)=vecF_(1)+vecF_(2)`
`=50,000 +(-10,000)`
`=40,000 N`
(ii) Mass of each wagon of the train = 2500 kg
Number of wagons of the train = 4
`therefore underset("(without engine)") ("Mass of the train(m)")=2500 xx 4=10,000 kg`
Net accelerating force acting on the train is
F = 40,000 N
From,
F = ma
`a=(F)/(m)=(40,000)/(10,000)=4m//s^(2)`
(iii) Mass of ONE wagon `(m_(1))=2500 kg`
Acceleration of the train `a=4m//s^(2)`
Force exerted by wagon-1 on wagon-2 is the accelerating force for the three rear wagons.

`therefore F'=3xx2500 xx 4m//s^(2)`
`=30,000N`
Discussion
3830.

Conductivity will increase with temperature in :

Answer»

metal
non-metal
semiconductor
None of these

Answer :C
Discussion
3831.

Two wires of same material of unchanged lengths in the ratio 2: 3 and diameters in the ratio 3 : 2 are stretched with equal forces . What is the ratio of their elongation produced ?

Answer»

a)`2 : 3`
B)`4:9`
C)`8:27`
d)`27:8`

ANSWER :C
Discussion
3832.

An electron is deflected by 1.76 mm in traversing a distance of 10 cm through an electric field of 1800 volt per metre perpendicular to its path. Find the electron charge to mass ratio, if average velocity of the electrons is 3xx10^(7) ms^(-1).

Answer»


Answer :`1.76 xx 10^(11)KG^(-1)`
Discussion
3833.

From which law and principle electric field by electric dipole can be obtained ?

Answer»

SOLUTION :The electric field of the PAIR of charges (- q and q) at any point in space can be found out from Coulomb.s law and the SUPERPOSITION principle.
The electric field at any GENERAL point P is obtained by adding the electric field E due to the charge - q and `E_(+q)` due to the charge q by the parallelogram law of vectors.
Discussion
3834.

The separation between the plates of a parallel-plate capacitor of capacitance 2muF is 2mm. The capacitor is initially uncharged. If a dielectric slab of surface area same as the capacitor, thickness 1mm, and dielectric constant 3 is introduced, and then the capacitor is charged to a potential difference difference 100V, the energy stored in the capacitor becomes _______mJ.

Answer»


Solution :Let the plate separation be d, the plate area be A, and the INITIAL capacitance be `C_(0)`.
Then, `C_(0)=(epsi_(0)A)/(d)`
After the dielectric SLAB is introduced, let the capacitance be C.
So, `(1)/(C)=(((d)/(2)))/(epsi_(0)A)+(((d)/(2)))/(3psi_(0)A) implies C=(3C_(0))/(2)`.
Since `C_(0)=2MUF, C=3muF`
So, energy stored in the capacitor, `U=(1)/(2)CV^(2)=(1)/(2)(3xx10^(-6))(100)^(2)=15mJ`.
Discussion
3835.

Angle of deviation for a thin prism of refractive index 1.5 is 4^(@) for an incident ray. If that prism is dipped in water, then for the same incident ray, angle of deviation would be ["given" " "mu_(water) = (4)/(3)]

Answer»

`1^(@)`
`2^(@)`
`8^(@)`
`16^(@)`

ANSWER :A
Discussion
3836.

The above is a plot of binding energy per nucleon E_(b), against the nuclear mass M, A,B,C,D,E,F correspond to different nuclei. Consider four reactions: (i) A+B to C+e (ii) C to A+B+epsi (iii) D+E to F+epsi and (iv) F to D+E+epsi where epsi is the energy released? In which reaction is epsi positive?

Answer»

(ii) and (iv)
(ii) and (iii)
(i) and (iv)
(i) and (iii)

Solution :In EQN (i) and (iv) energy released E is +ve. As two HIGHER unstable nuclei FUSE together to form a HEAVY unstable nuclei with release of energy and a heavy unstable nuclei splits into stable nuclei with release of energy.
Discussion
3837.

What did the doctor say?

Answer»

The boy WOULD die
He would remain a lifetime cripple
He would EARN a LOT of money
Both (A) and (B)

Answer :D
Discussion
3838.

Find how the volume density of the elastic deformation energy is distributed in a steel rod depending on the distance r from its axis. The length of the rod is equal to l, the torsion angle to varphi.

Answer»

SOLUTION :The ENERGY between radii r and `r+dr` is, by DIFFERENTIATION, `(pir^3dr)/(L)Gvarphi^2`
Its density is `(pir^3dr)/(2pirdrl)(Gvarphi^2)/(l)=1/2(Gvarphi^2r^2)/(l^2)`
Discussion
3839.

Intelset satellite works as a

Answer»

TRANSMITTER
RECEIVER
ABSORBER
repeater

Answer :D
Discussion
3840.

Which of these is not related to endoplasmic reticulum?

Answer»

It BEHAVES as transport channel for proteins between nucleus and cytoplasm
It TRANSPORTS materials between various REGIONS in cytoplasm
It can be the site of ENERGY generation
It can be the site for some biochemical activities of the cell

Answer :C
Discussion
3841.

A diffraction pattern is obtained using beam of red light. What happens if the red light is replaced by blue red

Answer»

no change
diffraction bands BECOME narrower and crowded together
bands become broader and father apart
bands disapear

Answer :B
Discussion
3842.

A toroid of non ferromagnetic has core of inner radius 25cm and outer radius 26cm. it has 3500 turns & carries a current of 1JA, then find the magnetic field at a point i) In the internal cavity of toroid ii) At the midpoint of the windings iii) At a point which is at a distance of 30cm from the centre of toroid

Answer»

Solution :i) B = 0
ii) `B=(mu_(0))/(2pi)(ni)/r=2xx10^(-7)xx(3500xx11)/(51xx10^(-2))xx2`
`=88/3xx10^(-3)=29.3xx10^(-3)T`
III) B = 0. Based on MAGNETISM for SOLENOID and toroid
Discussion
3843.

A body is falling freely under the action of force of gravity alone. No air resistance is there as duc laxly is falling in Vacuum. Out of the following quantities which of them remains contant during its fall?

Answer»

Kinetic ENERGY only
Potential energy only
Total mechanical energy
Total LINEAR MOMENTUM.

SOLUTION :Here it can be proved even mathematically that the sum of (P.E+K.E) is always constant when the body is falling freely in VACUM.
Discussion
3844.

Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively. (a) Give the signs of the potential difference V_(P)-V_(Q): V_(B)-V_(A). (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P, A and B. (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

Answer»

<P>

Solution :(a) As `V prop 1/r, V_(P) gt V_(Q)`, Thus, `(V_(P)-V_(Q))` is positive. Also `V_(B)` is less NEGATIVE than `V_(A)`. Thus `V_(B) gt V_(A)` or `(V_(B)-V_(A))` is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive.Similarly, `(P.E)_(A) gt (P.E)_(B)`and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric FIELD. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the KINETIC energy decreases in GOING from B to A.
Discussion
3845.

Large machine components that undergo prolonged, high-speed rotation are first examined for the possibility of failure in a spin test system. In this system, a component is spun up while inside a cylindrical arrangement of lead bricks and containment liner, all within a steel shell that is closed by a lid clamped into place. If the rotation causes the component to shatter, the soft lead bricks are supposed to catch the pieces for later analysis. In 1985, Test Devices, Inc. was spin testing a sample of a solid steel rotor of mass M = 272 kg and radius R = 38.0 cm. When the sample reached an angular speed omega of 14000 rev/min, the test engineers heard a dull thump from the test system, which was located one floor down and one room over from them. Investigating, they found that lead bricks had been thrown out in the hallway leading to the test room, a door to the room had been hurled into the adjacent parking lot, one lead brick had shot from the test site through the wall of a neighbor's kitchen, the structural beams of the test building had been damaged, the concrete floor beneath the spin chamber had been shoved downward by about 0.5 cm, and the 900 kg lid had been blown upward through the ceiling and had then crashed back onto the test equipment. The exploding pieces had not penetrated the room of the test engineers only by luck. How much energy was released in the explosion of the rotor?

Answer»

SOLUTION :The released energy was equal to the rotational kinetic energy K of the rotor just as it reached the angular SPEED of 1400 rev/min.

Calculations: We can find Eq. `(K=1//2Iomega^(2))`, but first we need an EXPRESSION for the rotational inertia I. Because the rotor was a disk that rotated like a merry-go-round, I is GIVEN in `(I=1//2MR^(2))`. Thus,
`I=1/2MR^(2)=1/2(272kg)(0.38m)^(2)=19.64kg*m^(2)`.
The angular speed of the rotor was
`omega=(14000"rev"//min)(2pi"rad"//"rev")((1min)/(60s))`
= `1.466xx10^(3)rad//s`.
Then, with we find the (huge) energy release :
`K=1/2Iomega^(2)=1/2(19.64kg*m^(2))(1.466xx10^(3)"rad"//s)^(2)`
= `2.1xx10^(7)J`.
Discussion
3846.

What would be the power of an engine required to lift 90 metric tonnes of coal per hour from a mine 200 m deep g=9.8 m/s^2.

Answer»

29 KW
39 KW
49 KW
60 KW

Answer :C
Discussion
3847.

The heavily and lightly doped regions of a bipolar junction transistor are respectively

Answer»

BASE and EMITTER
base and COLLECTOR
collector and base
emitter and base

ANSWER :D
Discussion
3848.

एक वैद्युत क्षेत्र विक्षेपित कर सकता है:

Answer»

एक्स-किरणों को
न्यूटॉनों को
एल्फा-कणो को
गामा-किरणों को

Answer :C
Discussion
3849.

In an LCR circuit , can the amplitude of the voltage across L be greater than that of an emf of the generator . The LCR circuit contains an AC source of emf (Generator emf)E = 10 V,R = 10 Omega ,L = 1H, C = 1mu F . Find the amplitude of the voltage across L at resonance condition .

Answer»

Solution :Voltage ACROSS ` L = V_L = IX_L = I omegaL`
At resosnance `omega = (1)/(sqrtLC)and I = E/R`
` therefore V_L = I. (1)/(sqrtLC).L = E/R sqrt(L/C) = 10/10 sqrt( (1)/(1 XX 10^(-6) ) ) = 1000 V`
this is MUCH larger than the generator emf E = 10 V . so the answer is yes .
Discussion
3850.

The number of electrons, neutrons and protons in a species are equal to 10,8 and 8 respectively. The proper symbol of the species is

Answer»

`""^(16)O_(8)`
`""^(18)O_(8)`
`""^(18)Ne_(10)`
`""^(16)O_(8)^(2-)`

Solution :Number of electrons =8+2=10
Number of PROTONS =8
Number of neutrons, N=8
ATOMIC number, Z=number of protons=8
Mass number, A+Z+N=8+8=16
The PROPER symbol of the SPECIES is `""^(16)O_(8)^(2-)`.
Discussion