Test: BITSAT Past Year Paper- 2010 - JEE MCQ

# Test: BITSAT Past Year Paper- 2010 - JEE MCQ

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## 150 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Test: BITSAT Past Year Paper- 2010

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Test: BITSAT Past Year Paper- 2010 - Question 1

### If P represents radiation pressure, c represents speed of light and Q represents radiation energy striking a unit area per second, the non-zero integers x, y and z such that PxQycz is dimensionless, are.

Test: BITSAT Past Year Paper- 2010 - Question 2

### The position x of a particle varies with time (t) as x = A t2 – B t3. The acceleration at time t of the particle will be equal to zero. What is the value of t?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 2

Given that x = A t2 – B t3
∴ velocity dx/dy = 2A t - 3 Bt 2
and  acceleration =   = 2 A - 6Bt
For acceleration to be zero 2A – 6Bt = 0.

Test: BITSAT Past Year Paper- 2010 - Question 3

### Two projectiles A and B are thrown with the same speed but angles are 40º and 50º with the horizontal. Then

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 3

lesser is the value of θ, lesser is sinq and hence lesser will be the time taken. Hence A will fall earlier.

Test: BITSAT Past Year Paper- 2010 - Question 4

A body is travelling in a circle at a constant speed. It

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 4

Body moves with constant speed it means that tangential acceleration aT = 0 & only centripetal acceleration aC exists whose direction is always towards the centre or inward (along the radius of the circle).

Test: BITSAT Past Year Paper- 2010 - Question 5

Two blocks are connected over a massless pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed. The mass of block B in kg is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 5

Considering the equilibrium of A, we get 10a = 10g sin 30º – T – mN
where N = 10g cos 30°

but a = 0, T = mBg

⇒ mB = 3.268 ≈ 3.3 kg

Test: BITSAT Past Year Paper- 2010 - Question 6

A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v1 and v2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 6

Applyin g law of conservation of lin ear momentum m1v1 + m2v2 = 0,

Test: BITSAT Past Year Paper- 2010 - Question 7

The potential energy for a force field is given by U (x, y) = cos (x + y). The force acting on a particle at position given by coordinates (0, π/4) is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 7

Test: BITSAT Past Year Paper- 2010 - Question 8

A long string is stretched by 2 cm and the potential energy is V. If the spring is stretched by 10 cm, its potential energy will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 8

Test: BITSAT Past Year Paper- 2010 - Question 9

The ratio of the accelerations for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of angle ‘θ’ without slipping and slipping down the incline without rolling is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 9

For solid sphere rolling without slipping on inclined plane, acceleration

For solid sphere slipping on inclined plane without rolling, acceleration
a2 = g sin θ
Therefore required ratio =

Test: BITSAT Past Year Paper- 2010 - Question 10

A system consists of three particles, each of mass m and located at (1, 1), (2, 2) and (3, 3). The co-ordinates of the centre of mass are

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 10

The coordinates of C.M of three particle are

so coordinates of C.M. of three particle are (2, 2) V = 300 m/s

Test: BITSAT Past Year Paper- 2010 - Question 11

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ‘R’ around the sun will be proportional to

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 11

F = KR- n = MRω2 ⇒ ω2 = KR- (n +1)

[where K' = K1/2, a constant]

Test: BITSAT Past Year Paper- 2010 - Question 12

Two planets A and B have the same material density. If the radius of A is twice that of B, then the ratio of the escape velocity vA/vB is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 12

as ve ∝ R for same density,

Test: BITSAT Past Year Paper- 2010 - Question 13

The upper end of a wire of diameter 12mm and length 1m is clamped and its other end is twisted through an angle of 30°. The angle of shear is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 13

Test: BITSAT Past Year Paper- 2010 - Question 14

A spherical ball is dropped in a long column of a viscous liquid. The speed (v) of the ball as a function of time (t) may be best represented by

Test: BITSAT Past Year Paper- 2010 - Question 15

Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 15

Surface energy of bigger drop,
E = 4πR2T = 4 x 22/3 πr2T = 28/3 πr2T

Test: BITSAT Past Year Paper- 2010 - Question 16

Two circular plates of radius 5 cm each, have a 0.01 mm thick water film between them. Then what will be the force required to separate these plate (S.T. of water = 73 dyne/cm) ?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 16

= 36.5 π ≈ 115 newton

Test: BITSAT Past Year Paper- 2010 - Question 17

One kilogram of ice at 0°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take specific heat of water = 4200 kJ/kg-°C, Latent heat of ice = 336 kJ/kg)

Test: BITSAT Past Year Paper- 2010 - Question 18

In the equation PVγ = constant, the value of γ is unity. Then the  process is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 18

PV = constant represents isothermal process.

Test: BITSAT Past Year Paper- 2010 - Question 19

An ideal refrigerator has a freezer at a temperature of 13ºC. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 19

T2 = 273 – 13 = 260,

or T1 – 260 = 52; T1 = 312 K,
T2 = 312 – 273 = 39°C

Test: BITSAT Past Year Paper- 2010 - Question 20

3 moles of an ideal gas at a temperature of 27°C are mixed with 2 moles of an ideal gas at a temperature 227°C, determine the equilibrium temperature of the mixture, assuming no loss of energy.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 20

Energy possessed by the ideal gas at 27°C is

Energy possessed by the ideal gas at 227°C is

If T be the equilibrium temperature, of the mixture, then its energy will be

Since, energy remains conserved,
Em = E+ E2

or T = 380 K  or 107°C

Test: BITSAT Past Year Paper- 2010 - Question 21

A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration 3 ms–2 is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 21

Test: BITSAT Past Year Paper- 2010 - Question 22

A wave y = a sin (ωt – kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 22

Equation of a wave
y1 = a sin (ωt – kx)         ....(i)
Let equations of another wave may be,
y2 = a sin (ωt + kx)         ....(ii)
y3 = –a sin (ωt + kx)      ....(iii)
If Eq. (i) propagate with Eq. (ii), we get y = 2a cos kx sin ωt
If Eq. (i), propagate with Eq. (iii), we get y = –2a sin kx cos ωt
At x = 0, y = 0, wave produce node
So, Eq.(iii) is the equation of unknown wave

Test: BITSAT Past Year Paper- 2010 - Question 23

A source of sound produces waves of wavelength 60 cm when it is stationary. If the speed of sound in air is 320 m s-1 and source moves with speed 20 m s-1, the wavelength of sound in the forward direction will be nearest to

Test: BITSAT Past Year Paper- 2010 - Question 24

A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked with the surface?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 24

The given square of side L may be considered as one of the faces of a cube with edge L. Then given charge q will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by
φ = q/ε0
Hence, electric flux through one face of the cube for the given square will be

Test: BITSAT Past Year Paper- 2010 - Question 25

Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10-2 C and 5 × 10-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 25

Test: BITSAT Past Year Paper- 2010 - Question 26

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 26

Test: BITSAT Past Year Paper- 2010 - Question 27

The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 27

...(i)

Substituting the values in equation (i)

Test: BITSAT Past Year Paper- 2010 - Question 28

Which of the following quantities do not change when a resistor connected to a battery is heated due to the current?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 28

Only number of free electrons is constant, other factors are temperature dependent.

Test: BITSAT Past Year Paper- 2010 - Question 29

The magnetic field at the origin due to the current flowing in the wire is –

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 29

Test: BITSAT Past Year Paper- 2010 - Question 30

The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 30

Test: BITSAT Past Year Paper- 2010 - Question 31

Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1 : 2.
The ratio of self-inductance of two solenoids is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 31

From

we get,

Test: BITSAT Past Year Paper- 2010 - Question 32

An alternating voltage V = V0 sin ωt is applied across a circuit. As a result, a current I = I0 sin (ωt – π/2) flows in it. The power consumed per cycle is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 32

The phase angle between voltage V and current I is π/2.

Test: BITSAT Past Year Paper- 2010 - Question 33

A resistance R and inductance L and a capacitor C all are connected in series with an AC supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find the potential difference across R, L and C.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 33

VR = iR = 5 × 16 = 80 Volt
VL = i × (ωL) = 5 × 24 = 120 Volt
VC = i × (l /ωC) = 5 × 12 = 60 Volt

Test: BITSAT Past Year Paper- 2010 - Question 34

The diameter of the objective of a telescope is a, its magnifying power is m and wavelength of light is λ, The resolving power of the telescope is:

Test: BITSAT Past Year Paper- 2010 - Question 35

The photoelectric threshold of metal is 2000Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1500Å is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 35

Test: BITSAT Past Year Paper- 2010 - Question 36

A material particle with a rest mass m0 is moving with a velocity of light c. Then the wavelength of the de Broglie wave associated with it is :

Test: BITSAT Past Year Paper- 2010 - Question 37

Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 37

Number of spectral lines =

Test: BITSAT Past Year Paper- 2010 - Question 38

Which of the following is best nuclear fuel

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 38

Pluton ium 239 is pr ocessed by br eeder mechanism to be used as nuclear feul.

Test: BITSAT Past Year Paper- 2010 - Question 39

A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 39

IC = IE –IB = 90 – 1 = 89 m A

Test: BITSAT Past Year Paper- 2010 - Question 40

A d.c. battery of V volt is connected to a series combination of a resistor R and an ideal diode D as shown in the figure below. The potential difference across R will be

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 40

In forward biasing, the diode conducts. For ideal junction diode, the forward resistance is zero; therefore, entire applied voltage occurs across external resistance R i.e., there occurs no potential drop, so potential across R is V in forward biased.

Test: BITSAT Past Year Paper- 2010 - Question 41

The vapour density of ozone is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 41

We know that,
Molecular weight of compound or molecules = 2 × V. D.
Vapour density (V. D.) of ozone molecules

Hence, V. D. of O3 is 24.

Test: BITSAT Past Year Paper- 2010 - Question 42

In redox reaction 1 g-eq of reducing agent requires P gm-eq. of oxidising agent. The value of P is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 42

In redox reaction,
g equivalent of reducing agent = g. equivalent of oxidising agent
Hence 1g equ. of reducing agent = P g equ. of oxidising agent.

Test: BITSAT Past Year Paper- 2010 - Question 43

Chloride ion and potassium ion are isoelectronic. Then :

Test: BITSAT Past Year Paper- 2010 - Question 44

Which of the following pairs has both members from the same period of periodic table :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 44

11Na ⇒ 2, 8, 1; 17Cl ⇒ 2, 8, 7
These have same number of shells. Hence, they are the elements of the same period.

Test: BITSAT Past Year Paper- 2010 - Question 45

In the periodic table, with the increase in atomic number, the metallic character of an element

Test: BITSAT Past Year Paper- 2010 - Question 46

Which of the following statements is/are true?
1. PH5 and BiCl5 do not exist.
2. pπ – dπ bond is present in SO2.
3. I3+ has bent geometry.
4. SeF4 and CH4 have same shape.

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 46

All the three statements are correct.

Test: BITSAT Past Year Paper- 2010 - Question 47

When the temperature is raised, the viscosity of liquid decreases this is because

Test: BITSAT Past Year Paper- 2010 - Question 48

At a constant volume the specific heat of a gas is 0.075 and its molecular weight is 40. The gas is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 48

⇒ CV = q × m
= 0.075 × 40
= 3.0 cal
CP – Cv = R
CP = CV + R
= 3 + 2 = 5

Monoatomic gas.

Test: BITSAT Past Year Paper- 2010 - Question 49

Which of these is least likely to act as Lewis base?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 49

BF3 is Lewis acid (e pair acceptor)

Test: BITSAT Past Year Paper- 2010 - Question 50

The Ksp of CuS, Ag2S and HgS are 10–31,10–44 and 10–54 respectively. The solubility of these sulphides are in the order :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 50

For CuS, solubility = (10–31)1/2;
for  HgS =

Test: BITSAT Past Year Paper- 2010 - Question 51

In which of the following reactions, H2O2 is acting as a reducing agent?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 51

SO2 changes to H2SO4 (O.N. changes from +4 to +6 oxidation)
2KI →  I2 (O.S. changes from –1 to 0 oxidation)
PbS → PbSO4 (O.S. changes from –2 to +6 oxidation)
Ag2O → 2Ag (O.S. changes from +1 to 0 reduction)

Test: BITSAT Past Year Paper- 2010 - Question 52

Sodium peroxide in contact with moist air turns white due to formation of :

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 52

Na2O2 + H2O → 2NaOH + 1/2 O2

Test: BITSAT Past Year Paper- 2010 - Question 53

Which of the following is similar to graphite

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 53

Graphite and boron nitride have similar structure.

Test: BITSAT Past Year Paper- 2010 - Question 54

The number of geometrical isomers of CH3 – CH = CH – CH = CH – CH = CHCl is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 54

The given structure has three double bonds whose each carbon atom is differently substituted hence number of geometrical isomers will be 2n = 23’ = 8, where n is the number of double bonds whose each carbon atom is differently substituted.

Test: BITSAT Past Year Paper- 2010 - Question 55

According to IUPAC system, the correct name of the compound having the formula

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 55

Test: BITSAT Past Year Paper- 2010 - Question 56

Liebig’s method is used for the estimation of

Test: BITSAT Past Year Paper- 2010 - Question 57

Hyperconjugation involves

Test: BITSAT Past Year Paper- 2010 - Question 58

Name of following reaction is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 58

The given reaction is Diel’s Alder reaction.

Test: BITSAT Past Year Paper- 2010 - Question 59

The unsaturated hydrocarbon which on ozonolysis gives one mole each of formaldehyde, acetaldehyde and methyl glyoxal (CH3COCHO) is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 59

Test: BITSAT Past Year Paper- 2010 - Question 60

Minamata disease is due to pollution of

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 60

Minamata is caused by Hg poisoning.

Test: BITSAT Past Year Paper- 2010 - Question 61

Phosphate pollution is caused by:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 61

Phosphate pollution is caused by sewage and agricultural fertilizers.

Test: BITSAT Past Year Paper- 2010 - Question 62

Eutrophication causes reduction in

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 62

Eutrophication causes reduction in dissolved oxygen.

Test: BITSAT Past Year Paper- 2010 - Question 63

Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at – 0.3°C. How much ethylene glycol must be added to 5 kg of water?
(Kf = 1.86 K Kg mol–1)

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 63

ΔTf = 0.3 ° C

∴ WB = 50 g
The amount used should be more than 50 g.

Test: BITSAT Past Year Paper- 2010 - Question 64

Which of the following will form the cathode with respect to iron anode in an electrolyte cell?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 64

In electrolytic cell the cathode is of higher reduction potential.

Test: BITSAT Past Year Paper- 2010 - Question 65

The activation energy for a simple chemical reaction A → B is Ea in forward direction. The activation energy for reverse reaction

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 65

Since the nature of reaction (i.e. exothermic or endothermic) is not given, Ea for reverse reaction can be more or less.

Test: BITSAT Past Year Paper- 2010 - Question 66

The following data are for the decomposition of ammonium nitrite in aqueous solution:

The order of reaction is:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 66

NH4 NO2 → N2 + 2H2O
Volume of N2 formed in successive five minutes are 2.75 cc, 2.40 cc and 2.25 cc which is in decreasing order. So rate of reaction is dependent on  concentration of NH4 NO2. As decrease is not very fast so it will be first order reaction.

Test: BITSAT Past Year Paper- 2010 - Question 67

Which liberates ammonia when treated with water?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 67

All nitrides react with H2O to give NH3 and CaCN2 also react with H2O
CaNCN + 3H2O → CaCO3 + NH3

Test: BITSAT Past Year Paper- 2010 - Question 68

The correct order of reactivity of halogens with alkalies is

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 68

Reactivity follows the order F > Cl > Br > I

Test: BITSAT Past Year Paper- 2010 - Question 69

In the manufacture of iron from haematite, limestone is added to act as:

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 69

Limestone (CaCO3) is mixed with Fe2O3 and it acts as flux to form slag (CaSiO3).

Test: BITSAT Past Year Paper- 2010 - Question 70

Which of the following has square planar geometry?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 70

[PtCl4]2– has square planar geometry. Pt : 5d96s1

Two electrons are removed from 5d shell and 6s shell. So, hybridisation takes place is dsp2 i.e. square planar geometry.

Test: BITSAT Past Year Paper- 2010 - Question 71

In which of the following conversions, phosphorus pentachloride is used as the reagent?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 71

When ethyl alcohol is treated with PCl 5 , then ethyl chloride is formed.
CH3CH2 - OH + PCl5 CH3CH2 - Cl + HCl + POCl3

Test: BITSAT Past Year Paper- 2010 - Question 72

Match List I (Reaction) with List II (Reagent) and select the correct answer using the codes given below the lists:

Test: BITSAT Past Year Paper- 2010 - Question 73

Which of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?

Detailed Solution for Test: BITSAT Past Year Paper- 2010 - Question 73

CH3OH does not have – CH(OH)CH3 group hence it will not form yellow precipitate  with an alkaline solution of iodine (haloform reaction).

Test: BITSAT Past Year Paper- 2010 - Question 74

Formic acid and acetic acid can be distinguished by: