SRMJEE Mock Test - 4 (Medical) - JEE MCQ

Steel is more elastic than rubber because its elasticity is more. Steel comes back to its original shape faster than rubber when deforming forces are removed.

Power = Work / Time
W = P × t → Explains the 1st and 2nd options
Energy = qΔV → Explains the 3rd option
Joule-second (J·s) is the unit of Planck’s constant, h.

For series combination,

= 12 k V

According to the question, we are adding impurity to the metal, then the number of electrons increases. If the number of electrons increase, their number of collision with the metal atoms also increase.
Electrical conductivity and thermal conductivity is due to movement of electrons from one side of conductor to other side. Increased number of collision, increases the thermal and electrical resistance.
So, electrical and thermal conductivities both decrease.

If width of the slits are different, the ray undergoing interference will be of different intensity. So, when they will be out of phase, they will not completely cancel each other, resulting in minimum non-zero intensity.

The conversion of RCOOH to RCH₂OH can be obtained by lithium aluminium hydride.
NaBH₄ does not reduce carboxylic acids.
Red P/HI would reduce it to the hydrocarbon stage.
Zn-Hg/conc. HCl: Clemmensen reduction is done for conversion of aldehydes to hydrocarbons

2N₂O₅ → 4 NO₂ + O₂

Production rate of O₂ =
Production rate of NO₂ =
Disappearance rate of N₂O₅ =

[Rate of disappearance of N₂O₅ = Twice the production rate of O₂]

Freundlich gave an empirical relationship between quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature.

(x is mass of gas adsorbed on a mass m of the adsorbent at a pressure p, k and n  are constant, which depends on nature of adsorbent and gas at a particular temperature)

From above equation of Freundlich adsorption isotherm

Taking log both sides 

Now if 1/n = 0, so the above equation will be independent of p and, we will get a straight line parallel to  x-axis as required. 

Due to the bigger size of colloidal particles the number of particles in a colloidal solution is comparatively small as compared to a true solution. Hence the values of colligative properties are of small order as compared to values shown by true solutions at same concentration.

Diazotization reaction of primary aliphatic amine in the presence of HNO₂ forms an unstable alkyl diazonium compound.

Which forms a carbocation after removal of dinitrogen gas. This carbocation undergoes hydrolysis reaction to give alcohol.

While rest other options are aromatic, amine and diazonium compound of aromatic amine is stable by resonance.

Reduction by NaBH₄ gives primary alcohol, With the help PBr₃ substitution reaction occurred and formed PhCH₂Br. In the presence of Li, Organolithium reagent will form and this will attack on oxirane ring, after the opening of oxirane ring, Ph-CH₂-CH₂-CH₂-OH product will form. 

The correct answer is Chlamydomonas.

• Chlamydomonas is found in stagnant water and on damp soil as well as freshwater, seawater, and even in snow.
• About Chlamydomonas-
  • Class: Chlorophyceae
  • Order: Chlamydomonadales
  • Scientific name: Chlamydomonas
  • Phylum: Chlorophyta
  • Higher classification: Chlamydomonadaceae

Viroids are unique infectious agents made up solely of a short strand of RNA, without a protein coat, known for causing diseases in plants such as potato spindle tuber disease. This makes Option C correct. Option A incorrectly describes prions, not viroids; Option B is false as viroids are smaller than viruses and lack DNA; Option D describes lichens, not viroids.

The characteristic feature of sponges (phylum Porifera) among the options provided is the presence of Ostia. Ostia are tiny pores or openings on the body of a sponge that allow water to enter its central cavity, called the spongocoel. These ostia are responsible for the filtration of water, which is a vital process for sponges to obtain food and oxygen.

Analyzing each statement:

Statement (a) is correct; mosses indeed have complex mechanisms for spore dispersal, utilizing various environmental factors.

Statement (b) is incorrect; in liverworts, the sporophyte develops from the zygote and is dependent on the gametophyte, not directly from spores.

Statement (c) is correct; Polytrichum can reproduce vegetatively through budding in the secondary protonema.

Statement (d) is incorrect; Marchantia, like most liverworts, is homosporous and produces only one type of spore.

Statement (e) is correct; sphagnum moss, or bog moss, contributes to the gradual filling of ponds and lakes by trapping sediments.

The incorrect statements are (b) "In liverworts, the haploid free living sporophyte is formed by spore germination," and (d) "Marchantia is a heterosporous bryophyte," making Option C the correct choice.

Cockroaches, like many other insects, possess an open circulatory system. This system is characterized by a limited network of blood vessels and a heart that pumps the hemolymph (blood) into the body cavity, where it directly bathes the organs. There is no distinction between blood and interstitial fluid in this type of system, unlike in closed circulatory systems.

(i) Diploblastic: Coelenterates (Cnidarians) are diploblastic as they have two germ layers: ectoderm and endoderm.

(ii) Triploblastic - Platyhelminthes to Chordates are triploblastic as they possess three germ layers: ectoderm, mesoderm, and endoderm.

(iii) Acoelomate - Platyhelminthes are classified as acoelomates.

(iv) Pseudocoelomate - Aschelminthes/Roundworms have a pseudocoelom.

(v) Eucoelomate - Annelids to Chordates have a true coelom.

Renal calculi are characterized by the presence of an insoluble mass of crystallized salts within the kidney. These formations can vary in size and composition, often leading to symptoms such as severe pain, hematuria (blood in urine), and urinary obstruction. It is crucial to manage these calculi through dietary changes, hydration, and medical interventions to prevent complications like kidney damage or recurrent stone formation.

Topic in NCERT: Renal calculi

Line in NCERT: "Renal calculi: Stone or insoluble mass of crystallised salts (oxalates, etc.) formed within the kidney."

A. Beta diversity - Between community diversity
B. Rich biodiversity - Tropical areas
C. Gamma diversity - Diversity of whole geographical region
D. Extinct species - Dodo
E. Critically endangered species - Great Indian bustard

Earth summit at Rio de Jenerio (1992), Brazil promoted Convention on Biological Diversity (CBD) which was signed by 152 nations. Its recommendations came into effect on 29th Dec. 1993. India became a party to this Convention on Biological Diversity in May, 1994.

Number of deaths is more than the number of births, showing a declining population.

RNAi or RNA interference takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA. Using Agrobacterium vectors, nematode specific genes are introduced into the host plant (tobacco plant). The introduction of DNA was such that it produced both sense and anti-sense RNA in the host cells. These two RNAs being complementary to each other formed a dsRNA (double stranded RNA) that initiated RNAi.

Let the total job be R.
Let the job done by Sonam in 1 day be O.
Let the job done by Simon in 1 day be I.
So, xO = R, or O = R/x ... (i)
And (x - 4)I = R or I = R/(x - 4) ... (ii)
Now, working together they take 5 days.
So, 5(I + O) = R ... (iii)
Plugging in the values of I and O from equations (i) and (ii) into equation (iii), equation (iii) becomes
5((R/(x - 4)) + R/x) = R
Or x² - 14x + 20 = 0
Thus, answer option 4 is correct.

The movements of Rohan are shown in the figure below:

Clearly, AD = BC = 2 km
So, required distance = AE
= (DE - AD)
= (3 - 2) km = 1 km

From the above diagram AF = 1 km.

We know that the area of a circle is given by:

Area = πr²

Let the radii of the two circles be r₁ and r₂.

According to the given ratio of areas:

πr₁² / πr₂² = 25 / 64

Canceling π from both sides:

r₁² / r₂² = 25 / 64

Taking the square root on both sides:

r₁ / r₂ = 5 / 8

The radius of the second circle is given as 16 m.

Substituting r₂ = 16:

r₁ / 16 = 5 / 8

Multiplying both sides by 16:

r₁ = 16 × (5/8) = 10 m

Thus, the correct answer is option B (10 m).

It is given that;

Selling price (SP) = 540 Rupees

Loss = 10%

We know that,

In the case of 10% profit gain,

SP =  600 × 110 / 100 = 660 Rupees

The line of passage, 'He returned to find that Pablo had completed the picture and it was so amazingly beautiful and lifelike that he gave his son his palette and brushes and never painted again' clearly depicts that Pablo had become an amazing artist so, his father gave up painting.
Hence, option C is the right answer.