This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Which of the following statements is/are True? I. Disputes related to election of a President are decided by the Supreme Court. II. Disputes related to vacancy in the electoral college are settled by the Election Commission. III. In case the election of a President is declared void by the Supreme Court, the acts performed by a President before the data of such decision of the court get invalidated. (a) I, II and III (b) I and III (c) I only (d) III only |
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Answer» The correct option: (c) I only |
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| 2. |
Pb and Sn are extracted from their chief ore by :A. carbon reduction and self reductionB. self reduction and carbon reductionC. electrolysis and self reductionD. self reduction and electrolysis |
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Answer» Correct Answer - B The chief ore of lead is galena (PbS_ and is extracted with the help of self reduction. At the same time the chief ore of tin is tin stone `(SnO_(2))` and is isolated from that by carbon reduction (smelting). |
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| 3. |
Pb and Sn are extracted from their chief ore byA. carbon reduction and self reduction respectivelyB. self reduction and carbon reduction respectivelyC. electrolysis and self reduction respectivelyD. self reduction nad electrolysis respectively |
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Answer» Correct Answer - B Tin can be extracted by carbon reduction method only where as lead can be extracted either by self reduction method or by carbon reduction method. |
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| 4. |
The angle of prism is 60° and angle of deviation is 30°. In the position of minimum deviation, the values of angle of incidence and angle of emergence are:(a) i = 45°; e = 50° (b) i = 30°; e = 45°(c) i = 45°; e = 45° (d) i = 30°; e = 30° |
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Answer» (c) In the position of minimum deviation i =e=A+δM /2 =60+30/2=45o |
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| 5. |
Critical angle of light passing from glass to water is minimum for(a) red colour (b) green colour(c) yellow colour (d) violet colour |
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Answer» (d) violet colour |
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| 6. |
In primary rainbow what is the order of colours observed from earth ?(a) Violet innermost, red outermost.(b) Red innermost, violet outermost.(c) Random.(d) White and dark alternatively |
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Answer» (a) Violet innermost, red outermost. |
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| 7. |
Yellow light is refracted through a prism producing minimum deviation. If i1 and i2 denote the angle of incidence and emergence for this prism, then(a) i1 = i2 (b) i1 > i2(c) i1 < i2 (d) i1 + i2 = 90º |
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Answer» (a) In the position of minimum deviation, i1 = i2 . |
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| 8. |
If i = angle of incidence and r = angle of refraction, thenthe ratio sini/sinrI. is a constant for a pair of media.II. is called refractive index of medium 2 with respect to medium 1.III. is called absolute refractive index of medium 2.IV. varies with temperature.(a) I, II and III (b) II, III and IV(c) I, II and IV (d) I, III and IV |
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Answer» (c) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. Remember that the angles of incidence (i) and refraction (r) are the angles that the incident and its refracted ray make with the normal, respectively. We have, sin i/sin r =n21 ...(i) where, n21 is a constant, called the refractive index of the medium 2 w.r.t. the medium 1. Eq. (i) is the well known Snell’s law of refraction. We note that n21 is a characteristic of the pair of media (and also depends on the wavelength of light) but is independent of the angle of incidence |
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| 9. |
A convex mirror is used to form the image of an object. Then which of the following statements is/are true?I. The image lies between the pole and the focusII. The image is diminished in sizeIII. The image is real(a) I only (b) II only(c) I and III (d) I and II |
| Answer» (d) The image formed by a convex mirror is always virtual. | |
| 10. |
In secondary rainbow what is the order of colours observed from earth?(a) Violet innermost, red outermost.(b) Red innermost, violet outer most.(c) Random.(d) White and dark alternatively |
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Answer» (b) Red innermost, violet outer most. |
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| 11. |
When light is refracted, which of the following does not change ?(a) Wavelength (b) Frequency(c) Velocity (d) Amplitude |
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Answer» (b) Frequency does not change on refraction EXPLANATION: When light is refracted into a medium its speed changes but the frequency remains the same because the energy associated with a photon of this light is proportional to the frequency of the light and it remains unchanged. Since the wavelength λ = c/, i.e. λ ∝ c. So the wavelength changes. Option (c) is true. |
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| 12. |
In case of reflection over spherical surface, which of these are correct ?I. Normal is taken as perpendicular of tangent at point of incidence.II. Perpendicular to incident ray which is perpendicular to plane of incident ray.III. Line joining centre of curvature of mirror with point of incidence.IV. Line joining centre of curvature and pole of curved surface.(a) I and II (b) I and III(c) II and III (d) II and IV |
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Answer» (b) Normal is perpendicular to the tangent to surface at the point of incidence i.e., the normal is along the radius, the line joining the centre of curvature of the mirror to the point of incidence. So, geometric centre of a spherical mirror is called its pole while that of a spherical lens is called its optical centre. The line joining the pole and the centre of curvature of the spherical mirror is known as the principal axis. In the case of spherical lenses, the pricipal axis is the line joining the optical centre with its principal focus. |
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| 13. |
The rays of different colours fail to converge at a point after going through a converging lens. This defect is called(a) spherical aberration (b) distortion(c) coma (d) chromatic aberration |
| Answer» (d) Since refractive index for different wave length of light is different. Hence the different colours of light forms images at different position. This phenomenon is called chromatic aberration. | |
| 14. |
If the light moving in a straight line bends by a small but fixed angle, it may be a case of(a) reflection (b) refraction(c) diffraction (d) both (a) & (b) |
| Answer» (d) It occurs in both reflection & refraction & the angle of bending rays is constant. | |
| 15. |
What causes chromatic aberration?(a) Marginal rays(b) Central rays(c) Difference in radii of curvature of its surfaces(d) Variation of focal length of lens with colour |
| Answer» (d) The cause of chromatic aberration is that lens focusses different colours at different points. | |
| 16. |
Chromatic aberration in a lens is caused by(a) reflection (b) interference(c) diffraction (d) dispersion |
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Answer» Chromatic aberration in a lens is caused by (d) dispersion |
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| 17. |
Match the increasing orders given in column I with the property given in column II `{:("Column-I","Column-II"),((A)Na^+ ltF^(-) ltO^(2-)ltN^(3-),(p)"Electronegativity"),((B)Li^(+)ltNa^(+)ltK^(+)ltRb^(+)ltCs^(+),(q)"Nuclear charge"),((C )"OltSltFltCl",(r)"Size"),((D)Cl^(-)ltK^(+)ltCa^(2+)ltSc^(2+),(s)"Electron affinity"),(,(t)"Ionisation energy"):}` |
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Answer» Correct Answer - A-r ; B-q,r ; C-s ; D-p,q,s,t (A)For isoelectronic species, the ionic size decrease with increase in nuclear charge (B)Number of atomic shells increases, ionic size increases. (C )Correct order,as Cl has less inter electronic repulsions than F due to bigger size of 3p-subshell (D)Oxidation state increases, the electronegativity increases.For isoelectronic species ionisation energy and electron affinity increases with increasing nuclear charge. |
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| 18. |
Which of the following is a natural polymer?A. NylonB. TeflonC. LinenD. Orlon |
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Answer» Correct Answer - C Natural polymers are those polymers which are found in plants and animals. Among the given polymers liner is natural polymer which is made from cellulose (polymer of glucose molecules) |
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| 19. |
When a mixture of manganese dioxide, potassium hydroxide and potassium chlorate is fused, the product obtained isA. `K_(2)SO_(4)`B. `K_(2)MNO_(3)`C. `K_(2)MnO_(4)`D. `KMnO_(4)` |
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Answer» Correct Answer - C When a mixture of maganese dixoide, potassium hydroxide and potassium chlorate is fused then potassium chlorate first decomposes to give potassium chloride and oxygen gas. The formed `O_(2)` gas then reacts with `MnO_(2)` and KOH to give `K_(2)MnO_(4)`. The equations for the above reaction can be written as: `2KCIO_(3) (s) underset(MnO_(2))overset(Delta)(to) 2KCI (s) + 3O_(2) (g)` `2MnO_(2) 4KOH + O_(2) to 2K_(2) MnO_(4) + 2H_(2) O` |
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| 20. |
`B_2H_6+NH_3to`Addition compound (X) `overset(450 K)to` Y+Z(g) In the above sequence Y and Z are respectively :A. Borazine and hydrogenB. Boron and hydrogenC. Boron nitride and hydrogenD. Borazine and hydrogen |
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Answer» Correct Answer - D `3B_2H_5+6NH_3 to underset([x])(3[BH_2(NH_3)_2][BH_4])overset(450 K)tounderset([y])(2B_3N_3H_3)+underset([z])(12H_2)` |
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| 21. |
Hydrogen peroxide in its reaction with `KIO_(4)` and `NH_(2)OH` respectively, is acting as aA. reducing agent,oxidising agentB. reducing agent, reducing agentC. oxidising agent,oxidising agentD. oxidising agent, reducing agent |
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Answer» Correct Answer - A `H_(2)O_(2)` reacts with `KIO_(4)` in the following manner: `overset(+7)(KIO_(4))+H_(2)O_(2)rArr overset(+5)(KIO_(3))+H_(2)O+O_(2)` on reaction of `KIO_(4)` with `H_(2)O_(2)`, oxidation state of `I` varies from `+7` to `+5` i.e., decreases. Thus `KIO_(4)` get reduced hence,`H_(2)O_(2)` is reducing agent here. `H_(2)O_(2)` reacts with `NH_(2)OH` in the following manner. `overset(-1)(NH_(2)OH)+H_(2)O_(2)rArr overset(+3)(N_(2)O_(3))+H_(2)O` In this reaction, oxidation state of `N` varies from `-1` to +3 i.e., increases, hence `H_(2)O_(2)` is acting on an oxidising agent here. |
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| 22. |
From the following statements regarding `H_(2)O_(2)`, choose the incorrect statements:A. It can act only as an oxidizing agentB. It decomposes on exposure to lightC. It has to be stored in plastic or wax lined glass bottles in darkD. It has to be kept away from dust |
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Answer» Correct Answer - A It can act as an oxidizing as well as reducing agent. |
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| 23. |
When boron is fused with potassium hydroxide which pair of species are formed ?A. `K_2O+B_2O_3`B. `KO_2B_2H_6`C. `K_3BO_3+H_2`D. `K_3BO_3+H_2O` |
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Answer» Correct Answer - C `2B(s)+6KOH(s)overset("Fuse")to2K_3BO_3+3H_2` |
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| 24. |
a. How does basic character of oxides and hydroxides vary down the group in alkali metals? Why? b. How does reducing power of elements vary in group `1`? |
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Answer» a. It goes on increasing down the group due to decrease in ionisation energy and increase in metallic character. b. Reducing power of `Li` is highest, `Na` is lowest, then it goes on increasing :`LigtCsgtRbgtKgtNa`. |
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| 25. |
Urnaium is isolated from its ore by dissolving it as `UO_2(NO_3)_2` and separating it as solid `UO_2(C_2O_4).x H_2O`A 1.0 g sample of ore on treatment with nitric acid yielded 1.48 g `UO_2(NO_3)_2` which on further treatment with 0.4 g `Na_2C_2O_4` yielded 1.23 g`UO_2(C_2O_4).xH_2O`.Determine weight percentage of uranium in the original sample of x. |
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Answer» Correct Answer - 3 Mass of uranium in the sample `1.48/394xx238=0.894 g` Now, `uderset(3.756)(UO_2(NO_3)_2)+underset(2.985)(Na_2C_2O_4)+xH_2Oto(UO_2(C_2O_4)xH_2Odarr + 2NaNO_3` Here `Na_2C_2O_4` is the limiting reagent, therefore , m mole of `UO_2(C_2O_4).xH_2O` formed is 2.985. `implies M(UO_2(C_2O_4)).xH_2O=1.23/2.985xx1000`=412=238+32+88+18x `implies x=54/18=3` |
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| 26. |
Consider the following statements and pick out the wrong one. (1)The solubility, thermal stability and the basic character of the hydroxides of alkaline earth metals increases from`Mg(OH)_2` to `Ba(OH)_2` (2)The dehydration of hydrated chlorides, bromides and iodides of Ca,Sr and Ba can be achieved on heating. (3)The chlorides of both beryllium and aluminium are soluble in organic solvents and are strong lewis acids.A. (1) and (2) onlyB. (1) and (3) onlyC. (1),(2) and (3) onlyD. None |
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Answer» Correct Answer - D (1)The soubility increases down the group because the change in lattice energy is more as compared to hydration energy.Thermal stability and the basic character both increase donw the group as metallic character increases. (2)It is correct statement. (3)Beryllium and aluminium are diagonally related.Chlorides of both are covalent in nature and thus are soluble in organic solvents. Chlorides of both are electron deficient and thus act as strong Lewis acids . |
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| 27. |
The number of visible lines when an electrons returns from the `5^(th)` orbit to ground state in the hydrogen spectrum is P. Then value of P is: |
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Answer» Correct Answer - 3 No of visible lines means Balmer series ltbgt Hence n-2=5-2=3 |
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| 28. |
Find the ratio between the wavelength of the most energetic' spectral lines in the Balmer and Paschen series of the hydrogen spectrum. |
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Answer» \(\frac1{\lambda}=R(\frac1{n_1^2}-\frac1{n_2^2})\) R → Rydburg constant Balmer series n1 = 2 n2 = \(\infty\) for highest energy n2 → \(\infty\) \(\frac1{\lambda_B}=R(\frac1{2^2}-\frac1{\infty})\) \(\frac1{\lambda_B}=R(\frac1{4})\) \(\frac1{\lambda_B}=R(\frac R{4})\) \({\lambda_B}=\frac4{R}\) for paschen series n1 = 3, n2 = \(\infty\) \(\frac1{\lambda_p}=R(\frac{1}{(3^2)}-\frac1{(\infty^2)})\) \(\frac1{\lambda_p}=\frac{R}{p}\) \(\lambda_p=\frac{q}R,\) \(\frac{\lambda_b}{\lambda_p}=\cfrac{\frac4R}{\frac{q}R}\) ⇒ \(\frac4R\times\frac{R}q\) ⇒ 4 : q |
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| 29. |
The total number of elements in which electron gain enthalpy is negative `Li, P, N, O^(-),S^(-),C,F,Cl,S,Ne` |
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Answer» Correct Answer - 4 `CO_(2), Mn_(2)O_(7),CrO_(3),SO_(2)` |
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| 30. |
No of possible metamers of given molecular formula `C_(5)H_(10)O` (carbonyl only) |
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Answer» Correct Answer - 4 ml `C_(2)H_(2)+2.5 O_(2) rarr2CO_(2)+H_(2)O(l)` `a " "2.5" a use "2a` `CO+1//2O_(2)rarrCO_(2)` `b " "b//2" use "b` ` a + b = 10` `a_(17)=15-((5)/(2)a+(b)/(2))+2a +b` `b-a=4` `a+b=10` 2b = 14 b = 7 a = 3 `Vo_(2) = 15 - ((15+7)/(2))= 4 ml` |
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| 31. |
10 ml of a mixture of `C_(2)H_(2)` and CO gas was exploded with 15ml of oxygen. The gases after the reaction had a volume of 17 ml at 298 K calculate volume of oxygen remained after treatment the resulting gas mixture with KOH. |
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Answer» Correct Answer - 6 mol of `SO_(2) (M) = (12)/(64)g" " , " mol of "SO_(3)(N) = (10)/(80)g" "because(4M)/(N)=((4xx12)/(64))/((10)/(80))=6` |
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| 32. |
Roasting of sulphides gives the gas X as a by product. This is a colourless gas ith choking smell of burnt sulphur and causes great damage to repiratory organs as a result of acid rain. Its aqueous solution is acidic, acts as reducing agent and its acid has never been isolated. The gas X is :-A. `SO_(3)`B. `H_(2)S`C. `SO_(2)`D. `CO_(2)` |
| Answer» Correct Answer - C | |
| 33. |
The number of acidic oxides oxides among the following are `CO_(2),Rb_(2)O,Mn_(2)O_(7),CrO_(3),Na_(2)O,MgO,Al_(2)O_(3),SO_(2)` |
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Answer» Correct Answer - 6 Infra red lines = Paschen + Brackets + Pfund = 6 |
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| 34. |
Blister copper is refined by stirring moltem impure metal with green logs of wood because such a wood liberated hydrocarbon gases like `(CH_(4))`. The process X is called _____ and the Metal contains impurity of Y is ____:A. X=cupellation Y=CuO_(2)`B. X= poling `Y=Cu_(2)O`C. X=poling `Y=CuO`D. X=cupellation `Y=CuO` |
| Answer» Correct Answer - B | |
| 35. |
The major role of fluorpar (`CaF_2` ) which added in small quantities in the electrolyte reduction of alumina dissolved in fused cryolite `(N_(3)AlF_(6)` ) isA. as a catalystB. to make the fused conducting mixtureC. to lower the fusion temperature of the molten massD. to reduce the rate of oxidation of the carbon at anode. |
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Answer» Correct Answer - C Impurity in solid decreases melting point . |
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| 36. |
How is `SO_2` an air pollutant? |
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Answer» `SO_2` acts as air pullutant due to : (i). `SO_2` is strongly irritating to the respiratory tract. (ii). At a concentration 5 ppm, causes throat and eye irritation, resulting in cough tears and redness in eyes. It causes breathlessness and affects larynx i.e., voice box. (iii). At 0.03 ppm concentration causes chlorosis in plants. When the plant is exposed for long time to `SO_2` it slows down the formation of chlorophyll resulting in injury to the leaves including loss of green colour. This is called chlorosis. (iv). `SO_2` dissolves in rain water and produces acid rain which damages building materials especially marble `CaCO_3`. `SO_3+H_2OtoH_2SO_3` `CaCO_3+H_2SO_3toCaSO_3+H_2O+CO_2` |
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| 37. |
Why is dioxygen a gas but sulphur a solid? |
| Answer» Oxygen due to its small size and high electronegativity form `ppi-ppi` multiple bonds. Hence oxygen exists as a dietomic `(O_2)` molecule. These molecules are held together by weak van der waals forces of attraction which can be easily overcome by collision of the molecules at room temperature `O_2` is a gas at room temperature. On the other hand sulphur due to its large size and low electronegativity, do not form `ppi-ppi` multiple bonds. Instead it prefers to form `S-S` single bonds. Further , because of stronger `S-S` bond than `O-O` bond, sulphur has a much greater tendency for catenation and exists as `S_8` molecule. Due to large molecular mass of `S_8` unit, the van der Waals forces of attraction holding `S_8` molecules together is very large and sulphur exist as solid at room temperature. | |
| 38. |
Knowing the electron gain enthalpy values for `O rarr O^(Ө)` and `O rarr O^(2-)` as `- 141 kJ mol^-1` and `+ 702 kJ mol^-1` respectively, how can you account for the formation of a large number of oxides having `O^(2-)` species and not `O^(Ө)` ? |
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Answer» Formation of `MO_2` and `MO` involves the following steps: `M_(g)overset(triangle_iH_l)toM^(o+)overset(triangle_iH_2)toM^(2+)` `Ooverset(triangle_(eg)H_l)toO^(ɵ)underset("Energy")overset("Lattice")toM^(2+)(O^(ɵ))_(2(s))`. `M^(2+)+O^(2-)underset("energy")overset("lattice")toM^(2+)O^(2-)` (i). `O^(2-)` has a stable noble gas configuration. (ii). Due to higher charge on `O^(2-)` than on `O^(ɵ)` ions, tha lattice energy released during the formation of oxides containing `O^(2-)` ion is much larger than during the formation of oxides containing `O^(ɵ)` ion. Hence formation of oxides containing `O^(2-)` species is energetically more favourable than oxides containing `O^(ɵ)` species. |
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| 39. |
The increasing order of reactivity among group `1` elements is `LiltNaltKltRbltCs` whereas that among group 17 elements is `FgtClgtBrgtI`. Explain. |
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Answer» The elements of group `1` have only one electron in their respective valence shells thus have a strong tendency to lose this electron. The tendency to lose electrons, in turn, depends upon the `IE`. Since the `IE` decreases down the group, therefore the reactivity of group `1` elements increases in the same order:`LiltNaltKltRbltCs`. In contrast, the elements of group `17` have seven electrons in their respective valence shells and thus have a strong tendecy to accept one more electron. The tendency to accept electrons, in turn depends upon their electrode potentials. Since the electrode potentials of group `17` elements decrease in the order: `F(+2.87 V)gtCl(+1.36V)`,`Br(1.08 V)` and `I(+0.53 V)`, therefore, their reactivities also decrease in the same order:`FgtClgtBrgtI`. |
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| 40. |
Which of the following molecule is planer.A. `XeOF_(4)`B. `XeO_(4)`C. `XeF_(4)`D. `XeO_(3)` |
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Answer» Correct Answer - C Which of the ………. `XeF_(4)` is a planar molecular. |
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| 41. |
A photon has an energy of 4.00 x 10-19 J. Find the frequency. |
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Answer» Given \(E = 4 \times 10^{-19} J\) Frequency \(V =\, ?\) \(E = hV\) \(V = \frac Eh\) \(V = \frac{4\times 10^{-19}}{6.64 \times 10^{-34}}\) \(V = 0.602 \times 10^{15}\) \(V = 6.02 \times 10^{14} Hz\) |
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| 42. |
Adulteration in food product causes-(a) nutrition (b) malnutrition (c) low nutrition (d) healthy life |
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Answer» Adulteration in food product causes malnutrition. |
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| 43. |
Why laundry is an art and science? Give any four differences between household and commercial laundry. |
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Answer» Laundry is an art and science - It is science as it is based on application of scientific principles and techniques. It is art since application requires the mastering of certain skills to produce aesthetically pleasing results. Any other Four differences between household and commercial laundry
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| 44. |
To save clothes from insects.............. are applied.(A) Newspaper's ink(B) Dried neem leaves(C) Naphthalene balls (D) all the these. |
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Answer» To save clothes from insects Newspaper's ink, Naphthalene balls, Dried neem leaves are applied. |
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| 45. |
Why is blue applied to white clothes? Present four steps of applying blue to the clothes. OR Why is starch applied to clothes? Present four steps of applying starch to the clothes. |
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Answer» Blue is applied to white clothes because- It acts as a fabric whitener/brightener Four steps of applying blue to the clothes- 1.Add blue to water and dissolve well 2.Wet the article to be blued in water and squeezed properly 3.The article is opened and then dipped in the above blue solution 4.The article is squeezed and put in the sun OR Starch is applied to clothes because- It gives crispness/stiffness/brightness to the fabric Four steps of applying starch to the clothes- 1.Mix starch paste in water and dissolve well 2.Wet the article to be starched in water, squeeze it properly 3.Open the article and dip in starch solution 4.Squeeze and spread in the sun |
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| 46. |
Enlist four ways each to create variety and save time and money while planning a menu. |
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Answer» Four ways to create variety are- 1. Selection of foods from different food groups 2. Variety in color combination Blending of different colors makes food attractive, eye appealing and also enhances appetite. 3. Variety in texture The Texture in food refers to its softness, crispness and the solid or liquid state of the food. 4. Variety in taste and flavor A meal will tempt only if it has variety of flavor and aroma. 5. Any other Four ways to save time and money 1. Once the meals are planned, prepare a list of items and they can be purchased together. 2. Ingredients in bulk quantities cost less but should be purchased only when there is proper Provision for its storage. 3. Purchasing should be made at time periods when the markets are not very crowded. 4. Labor saving kitchen devices like mixer, fridge, pressure cooker, etc. should be used. 5. Any other |
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| 47. |
One summer evening ,Mr. Verma wore a starched cotton kurta for a party . He accidentally dropped water on his kurta. After returning home, he immediately folded and kept in the cupboard. After few months he was shocked to notice foul smell, patches and holes in this kurta.(i) What precaution Mr. Verma should have taken while storing his kurta? a. Should be destarched and dried in air b. Should be starched and ironed c. Stored in damp place d. Should not wash (ii) There was foul smell from kurta because of-a. Carpet Beetle b. Ants c.Mildew d. Moth (iii) Mr. Verma accidently put excessive blue while laundering his white cotton Kurta .Name the chemical which he can use to remove excessive blue from white cotton kurta-a. Vinegar b. Oil c. Hydrogen peroxide d. Sodium hypochlorite (iv) If heavily starched cotton kurta is stored, it can be attacked by ___________ . a. Silver fish b. Mildewc. Crickets d. Insects |
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Answer» (i) a. Should be destarched and dried in air (ii) c. Mildew (iii) a. Vinegar (iv) a. Silver fish |
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| 48. |
You are purchasing fruits from a street vendor. In what six ways you have to remain careful so that he does not cheat you while weighing the fruits? |
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Answer» Six ways to remain careful so that fruit vendor does not cheat while weighing the fruits- 1. Hollow bottom iron weights should not be used 2. Check magnet should not be there under the pan of the weighing scale 3. Pointers must be there in weighing scale 4. Vendor should not use irregular weights 5. Iron rings should not be hung on one side of weighing scale 6. Weighing balance must remain horizontal |
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| 49. |
Give reason-a) Cotton shirt should be de -starched before storing. b) Food stains should be immediately removed from woolen sweater. |
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Answer» a) Cotton shirt should be de -starched before storing- As starched clothes can be attacked by silverfish b) Food stains should be immediately removed from woollen sweater- As micro-organisms attack any food stain |
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| 50. |
a) What four points you should keep in mind while planning menu for 15-year-old boy? b) Suggest two ways of preparing a low-cost meal. |
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Answer» a) Four points to be kept in mind while planning menu for 15- year-old boy- 1. Menu should be nutritious 2. Easy to eat 3. Avoid over eating and dieting 4. Food should provide satiety b) Two ways of preparing a low-cost meal- 1. Prefer using seasonal fruits and vegetables 2. Substitute expensive food with low-cost nutritious food-For example use peanuts instead almonds 3. Use locally available food products |
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