SRMJEE Mock Test - 9 (Medical) - JEE MCQ

Ampere is the unit of current, so we can write it as Q/t
Hour is the unit of time, so we can write it as t.
Ampere-hour is = Q. Hence, ampere-hour is the unit of electric charge.

The fringe width in Young’s double-slit experiment

β = λD / d

where λ is the wavelength of light used
D is the distance between slit and screen.
d is the distance between the slit.

∴ β' = λ(2D) / (d/2) = 4λD / d = 4β

R_C = 2 kΩ and V₀ = 4 V
I_C = = 2 mA
β = I_C/I_B = 100
⇒ I_B = I_C/100 = 2 × 10^–5 A
V_in = I_BR_i = 2 × 10^–5 × 1 kΩ = 20 mV
Hence, the input signal voltage is 20 mV.

We know that,
ΔE = hc/λ ...(1)

K_α is the production of X-ray by transition from n = 2 to n = 1.
K_β is the production of X-ray by transition from n = 3 to n = 1.

As, E_Kβ > EK_α,
From equation (1),
λ_Kα > λ_Kβ.

So, λ = 0.7 Å for K_α.

The coefficient of viscosity is the ratio of tangential stress on the top surface of the film (exerted by block) to that of velocity gradient (vertically downwards) of the film. Since mass m moves with constant velocity, the string exerts a force equal to mg.

on the plate towards right. Hence, oil shall exert tangential force mg on the plate towards left.

=2.5 dyne s cm⁻²

Since the amine (X) reacts with benzenesulphonyl chloride and forms a product which is soluble in KOH, so amine (X) must be a primary amine.
Secondary amine forms a product which is insoluble in KOH.
Tertiary amine does not react with benzenesulphonyl chloride.

HCHO contain no α -H in its structure. Thus, in presence of NaOH it show Cannizaro reaction.

Heroin is an acyl derivative of morphine. It is also known as diacetylmorphine. It is 3,6-diacetyl derivative of morphine.

Rate of S_N1 reaction depends on the stability of carbocation (intermediate)
in molecule P after losing Cl⁻ carbocation will form and that will be stabilized by resonance with a double bond and electron-donating group −OCH₃
in molecule Q also carbocation will be stabilized by resonance with a double bond and −OCH₃ group but resonance with both is not continuous hence Q will be less reactive then P.
In R molecule there will not be any substitution reaction because of partial double bond character in C-Cl bond due to resonance with lone pair of Cl and a double bond, hence least reactive.
Order of reactivity is P > Q > R

Biodegradable polymers are a special class of polymer that breaks down after its intended purpose by bacterial decomposition process to result in natural by products such as gases, water, biomass, and inorganic salts
Poly β-hydroxybutyrate – co-β-hydroxy valerate (PHBV) is a biodegradable polymer, It is obtained by the condensation polymerisation of 3-hydroxybutanoic acid and 3 - hydroxypentanoic acid.  PHBV undergoes bacterial degradation in the environment.

The size of an species decreases with increasing nuclear charge because the attraction for the electrons increases. Thus, Al³⁺ is smaller in size.

This problem involves conceptual mixing of preparation of styrene and addition polymerisation of styrene. Students are advised to follow these steps.

• Complete the reaction first using information provided in question. (use elimination reaction)
• Then, using the product of above reaction as starting material identify the correct product.

Preparation of styrene

Styrene is obtained due to elimination reaction of 1-chloro-2 phenyl ethane in presence of strong base KO But
Polymerisation of styrene

From kjeldahl's method,
Percent of nitrogen = 1.4 x N x V / W
= 1.4 x 0.5 x 2 x 10 / 0.25
= 56%

Zn : Because of its maximum standard oxidation potential

Pericycle cells are thick-walled parenchymatous cells that lie next to the endodermis. Initiation of lateral roots and vascular cambium take place from these cells.

The atrial wall of our heart secretes a very important peptide hormone called ANF. The juxtaglomerular cells of kidney produce a peptide hormone called erythropoietin. The gastro-intestinal tract secrete peptide hormone called CCK

Depolarization of the axonal membrane happens due to an increase in Na⁺ permeability, which allows Na⁺ ions to rapidly enter the axoplasm, making Statement A correct. Statement B is also correct, as the diffusion of K⁺ out of the membrane restores the resting potential. Statement C is incorrect because the sodium-potassium pump moves 3 Na⁺ out for every 2 K⁺ in, not the other way around. Hence, the correct statements are A and B.
Topic in NCERT: Generation and Conduction of Nerve Impulse.
Line in NCERT: "The rise in the stimulus-induced permeability to Na* is extremely short-lived. It is quickly followed by a rise in permeability to K*. Within a fraction of a second, K+ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation."

A- Lactobacillus B- Trichoderma polysporum C-Yeast D-Penicillin

• The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece.
• This step is known as elution.

Line in NCERT: "The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution."

Glycocalyx (mucilage sheath) is the outermost layer of the bacterial cell envelope which consists of non-cellulosic polysacharides with or without proteins. It may occur in the form of loose sheath, called slime layer. If it is a thick coering, it is called capsule. Glycocalyx gives sticky character to the cell and is not absolutely essential for survival of bacteria. It prevents desiccation, protects from phagocytes, toxic chemicals and viruses and serves in attachment. It may give selective advantage though in certain situations.

Line in NCERT: "Glycocalyx differs in composition and thickness among different bacteria. It could be a loose sheath called the slime layer in some, while in others it may be thick and tough, called the capsule."

First 10 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Sum of first 10 odd natural numbers = 100
Mean = 100/10 = 10

The profit percentage is calculated using the formula:

Profit % = (SP - CP) / CP × 100

where SP is the total selling price and CP is the total cost price for the TV set as well as the DVD player.

Given Data:
Selling Price (SP) = ₹26,400
Cost Price (CP) = ₹18,000 + ₹4,000 = ₹22,000
Calculation:
Profit % = (26,400 - 22,000) / 22,000 × 100
= 4,400 / 22,000 × 100
= 20%

Conclusion: The profit percentage is 20%.

Given as
cosec(θ) − sin(θ) = m and sec(θ) – cos(θ) = n
Now,
m = cosecθ − sinθ
= 
n = secθ – cosθ

Required -  

Therefore option a will be correct option.