Which one of the following is NOT an advantage of inbreeding?
Inbreeding INCREASES (not decreases) homozygosity — so statement (A) is factually wrong, hence not an advantage.
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Which one of the following is NOT an advantage of inbreeding?
Inbreeding INCREASES (not decreases) homozygosity — so statement (A) is factually wrong, hence not an advantage.
Match List I with List II:
| List I | List II |
|---|---|
| A. Mast cells | I. Ciliated epithelium |
| B. Inner surface of bronchiole | II. Areolar connective tissue |
| C. Blood | III. Cuboidal epithelium |
| D. Tubular parts of nephron | IV. specialised connective tissue |
Mast cells → areolar CT; bronchiole → ciliated epi; blood → specialised CT; nephron tubule → cuboidal epi.
Statement I: During G₀ phase of cell cycle, the cell is metabolically inactive. Statement II: The centrosome undergoes duplication during S phase of interphase.
G₀ cells are metabolically ACTIVE but not dividing. Centrosome doubles in S phase.
Which of the following are NOT under the control of thyroid hormone?
A. Maintenance of water and electrolyte balance B. Regulation of basal metabolic rate C. Normal rhythm of sleep-wake cycle D. Development of immune system E. Support the process of R.B.Cs formation
Sleep-wake cycle is controlled by melatonin (pineal). Immune development is by thymus, not thyroid.
Select the correct statements with reference to chordates.
A. Presence of a mid-dorsal, solid and double nerve cord. B. Presence of closed circulatory system. C. Presence of paired pharyngeal gill slits. D. Presence of dorsal heart. E. Triploblastic pseudocoelomate animals.
Chordates: dorsal HOLLOW SINGLE nerve cord (A wrong), ventral heart (D wrong), coelomate (E wrong). B and C correct.
Match List I with List II.
| List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelengths (nm)) |
|---|---|
| A. $n_2 = 3$ to $n_1 = 2$ | I. 410.2 |
| B. $n_2 = 4$ to $n_1 = 2$ | II. 434.1 |
| C. $n_2 = 5$ to $n_1 = 2$ | III. 656.3 |
| D. $n_2 = 6$ to $n_1 = 2$ | IV. 486.1 |
Choose the correct answer from the options given below:
Balmer series: $3\to 2$: $H_\alpha = 656.3$ nm; $4\to 2$: $486.1$ nm; $5\to 2$: $434.1$ nm; $6\to 2$: $410.2$ nm.
Match List-I with List-II.
| List-I (Material) | List-II (Susceptibility ($\chi$)) |
|---|---|
| A. Diamagnetic | I. $\chi = 0$ |
| B. Ferromagnetic | II. $0 > \chi \geq -1$ |
| C. Paramagnetic | III. $\chi \gg 1$ |
| D. Non-magnetic | IV. $0 < \chi < \varepsilon$ (a small positive number) |
Choose the correct answer from the options given below:
Diamagnetic: $\chi$ slightly negative; Ferromagnetic: $\chi \gg 1$; Paramagnetic: small positive; Non-magnetic: $\chi = 0$.
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}\ \text{kg m}^2$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5}\ \text{Am}^2$; then the value of '$x$' is:
$T = 5/20 = 0.25$ s; $T = 2\pi\sqrt{I/MB}$ gives $M = 4\pi^2 I/(T^2 B) = 1280\pi^2 \times 10^{-5}$ Am².
An unpolarised light beam strikes a glass surface at Brewster's angle. Then
At Brewster's angle the reflected ray is fully plane-polarised; the refracted ray is only partially polarised.
Consider the following statements A and B and identify the correct answer:
A. For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
B. In a reverse biased $pn$ junction diode, the current measured in ($\mu A$), is due to majority charge carriers.
A is correct; B is wrong — reverse-bias current is from minority carriers.
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