BITSAT 2026 Question Paper with Solutions

The number of radial nodes for any orbital is given by the formula

For a $3𝑑$-orbital: $𝑛=3,𝑙=2$

Radial nodes $=3-2-1=0$

For a $4𝑝$-orbital: $𝑛=4,𝑙=1$

Radial nodes $=4-1-1=2$

Hence, the values are 0 and 2 respectively.

From standard gas behaviour charts (Compressibility factor Z vs $𝑝$ )

Curve $𝐴$ represents hydrogen $\left(H_2\right)$ or helium ( He ) because they show positive deviation $(𝑍>1)$ at all pressures due to negligible attractive forces.

Among the given gases, ${\mathrm{CO}}_2$ (curve O) stays closest to the $Z=1$ ideal line over a high-pressure range before increasing, whereas gases like $N_2\left(𝐵\right)$ and ${\mathrm{CH}}_4\left(𝐶\right)$ do not show ideal behaviour over a high pressure or compared or ${\mathrm{CO}}_2\left(𝐷\right)$.

Thus, ${\mathrm{CO}}_2$ shows maximum ideal behaviour at high pressure and curve $𝐴$ represents $H_2$.

When ferrous oxalate $\left({\mathrm{FeC}}_2{O}_4\right)$ acts as a reducing agent in an acidic medium, both the ${\mathrm{Fe}}^{2+}$ and the oxalate ion $\left(C_2{O}_{2-}^4\right)$ undergo oxidation.

Oxidation of iron ${\mathrm{Fe}}^{2+}⟶{\mathrm{Fe}}^{3+}+1{𝑒}^{-}$

Oxidation of oxalate $C_2{O}_{2-}^4⟶2{\mathrm{CO}}_2+2{𝑒}^{-}$

Total number of electrons lost per mole of ${\mathrm{FeC}}_2{O}_4$ is $1+2=3$. Therefore, its $𝑛$-factor is 3 .

Depression in freezing point is given by

                  $\Delta {𝑇}_{𝑓}=𝑖\cdot {𝐾}_{𝑓}\cdot 𝑚$

For acetic acid It dissociates into 2 ions $(𝑛=2)$ with degree of dissociation $𝛼=0.1$

        ${𝑖}_1=1+𝛼(𝑛-1)=1+0.1(2-1)=1.1$

For solute $𝑿$ It dimerises completely ( $𝑛=2$ associating, $𝛼=1$ ).

        $\begin{array}{cc}{𝑖}_2& =1+𝛼(\frac{1}{𝑛}-1)\\ & =1+1(\frac{1}{2}-1)=0.5\end{array}$

Since, the molalities are the same (0.1 m), the ratio of $\Delta {𝑇}_{𝑓}$ is just the ratio of their van't Hoff factors:

$\frac{\Delta {𝑇}_{𝑓\left(\text{ Acetic acid }\right)}}{\Delta {𝑇}_{𝑓\left(𝑥\right)}}=\frac{1.1}{0.5}=1.1:0.5$

In an fcc unit cell of edge length $𝑎$.

An octahedral void is present at the body center coordinates $(\frac{𝑎}{2},\frac{𝑎}{2},\frac{𝑎}{2})$.

Tetrahedral voids are located along the body diagonals at a distance of $\frac{\sqrt{\surd 3}𝑎}{4}$ from the corners, e.g., at coordinates $(\frac{𝑎}{4},\frac{𝑎}{4},\frac{𝑎}{4})$.

The shortest distance between the body-center octahedral void and a tetrahedral void is exactly one-fourth of the body diagonal.

Distance $=\sqrt{\surd {(\frac{𝑎}{2}-\frac{𝑎}{4})}^2\times 3}=\sqrt{\surd 3{\left(\frac{𝑎}{4}\right)}^2}=\frac{\sqrt{\surd 3}𝑎}{4}$

For a first-order reaction, the integrated rate law is

$𝑡=\frac{2.303}{𝑘}\log \left(\frac{𝑎}{𝑎-𝑥}\right)$

Half-life $\left({𝑡}_{1/2}\right)=\frac{0.693}{𝑘}$

Time for $99.9\%$ completion ( ${𝑡}_{99.9\%}$ ) :

$\begin{array}{cc}{𝑡}_{99.9\%}& =\frac{2.303}{𝑘}\log \left(\frac{100}{100-99.9}\right)\\ & =\frac{2.303}{𝑘}\log \left(1000\right)\\ & =\frac{2.303\times 3}{𝑘}=\frac{6.909}{𝑘}\end{array}$

Taking the ratio :

                          $\frac{{𝑡}_{99.9\%}}{{𝑡}_{1/2}}=\frac{\frac{6.909}{𝑘}}{\frac{0.693}{𝑘}}\approx 10$

In an Ellingham diagram, the metal whose oxidation line lies lower (more negative $\Delta {𝐺}^{\circ }$ ) can reduce the oxide of the metal whose line lies above it.

At temperature ${𝑇}_1$, the line for the oxidation of $𝑋(2𝑋+O_2⟶2𝑋O)$ is lower than line for the oxidation of $𝑀(2𝑀+O_2⟶2𝑀O)$.

Therefore, Metal $𝑋$ has a greater affinity for oxygen at ${𝑇}_1$ and can successfully reduce $𝑀O$ to metal $𝑀$.

In the complex ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$, cobalt is in $𝑎+3$ oxidation state, meaning its electron configuration is $3{𝑑}^6$.

Ammonia $\left({\mathrm{NH}}_3\right)$ acts as a strong field ligand for ${\mathrm{Co}}^{3+}$, leading to the pairing of electrons and forming a low-spin complex.

All 6 electrons occupy the lower energy ${𝑡}_{2𝑔}$ orbitals: ${\left({𝑡}_{2𝑔}\right)}^6{\left({𝑒}_{𝑔}\right)}^0$.

$\mathrm{CFSE}=[-0.4\times ($ number of ${𝑡}_{2𝑔}$ electrons $)+0.6\times$ (number of ${𝑒}_{𝑔}$ electrons)] ${\Delta }_{𝑜}$

$\mathrm{CFSE}=[-0.4\times 6+0.6\times 0]{\Delta }_{𝑜}=-2.4{\Delta }_{𝑜}$

For a weak monoprotic acid $\left(𝐻𝐴\right)$ titrated with a strong base $\left(\mathrm{NaOH}\right)$, the half-equivalence point occurs when half of the weak acid has been neutralised to form its conjugate base ( ${𝐴}^{-}$).

At the half-equivalence point, $\left[𝐻𝐴\right]=\left[{𝐴}^{-}\right]$. Using the Henderson-Hasselbalch equation

$\mathrm{pH}=p{𝐾}_{𝑎}+\log \left(\frac{\left[{𝐴}^{-}\right]}{\left[𝐻𝐴\right]}\right)\Rightarrow \mathrm{pH}=p{𝐾}_{𝑎}+\log \left(1\right)$

$\Rightarrow \mathrm{pH}={\mathrm{pK}}_{𝑎}$

Given, $\mathrm{pH}=4.0$, it follows that ${\mathrm{pK}}_{𝑎}=4.0$, which means ${𝐾}_{𝑎}={10}^{-4}=1.0\times {10}^{-4}$.

The dissolution reaction is

$\mathrm{AgCl}\left(𝑠\right)+2{\mathrm{NH}}_3\left(𝑎𝑞\right)\rightleftharpoons {[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2]}^{+}+{\mathrm{Cl}}^{-}\left(𝑎𝑞\right)$

The equilibrium constant for this combined reaction is

$\begin{array}{cc}& {𝐾}_{\mathrm{eq}}={𝐾}_{\mathrm{sp}}\times {𝐾}_{𝑓}.\\ & {𝐾}_{\mathrm{eq}}=(1.0\times {10}^{-10})\times (1.0\times {10}^7)=1.0\times {10}^{-3}\end{array}$

Let $𝑆$ be the solubility. Then $\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{+}^2\right]=𝑆$ and $\left[{\mathrm{Cl}}^{-}\right]=𝑆$.

The concentration of ${\mathrm{NH}}_3$ remaining is approximately $1.0-2𝑆\approx 1.0M$.

$\begin{array}{cc}{𝐾}_{\mathrm{eq}}& =\frac{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_{+}^2\right]{\mathrm{Cl}}^{-}}{{\left[{\mathrm{NH}}_3\right]}^2}\Rightarrow {10}^{-3}=\frac{𝑆\cdot S}{(1.0{)}^2}={𝑆}^2\\ 𝑆& =\sqrt{\surd {10}^{-3}}=\sqrt{\surd 10\times {10}^{-4}}\approx 3.16\times {10}^{-2}M\end{array}$

Peroxodisulphuric acid $\left(H_2{S}_2{O}_8\right)$, also known as Marshall's acid, possesses a peroxide bond $(-O-O-)$ directly linking the two sulphur-containing groups.

When potassium permanganate $\left({\mathrm{KMnO}}_4\right)$ is heated strongly, it undergoes thermal decomposition to produce potassium manganate $\left(K_2{\mathrm{MnO}}_4\right)$, manganese dioxide ( ${\mathrm{MnO}}_2$ ), and oxygen gas ( $O_2$ ). The balanced chemical equation is

$2{\mathrm{KMnO}}_4\stackrel{\Delta }{\to \leftarrow }{K}_2{\mathrm{MnO}}_4+{\mathrm{MnO}}_2+O_2$

Both ${\mathrm{NH}}_3$ and ${\mathrm{NF}}_3$ have pyramidal geometry. However, their dipole moments differ significantly due to bond polarities.

In ${\mathrm{NH}}_3$, nitrogen is more electronegative than hydrogen. The $N-H$ bond dipoles point towards nitrogen, which aligns with the orbital dipole of the lone pair, leading to a high net dipole moment.

In ${\mathrm{NF}}_3$, fluorine is more electronegative than nitrogen. The N-F bond dipoles point away from nitrogen, opposing the direction of the lone pair's orbital dipole. They partially cancel out, resulting in a much lower net dipole moment.

Thus, $𝜇\left({\mathrm{NH}}_3\right)>𝜇\left({\mathrm{NF}}_3\right)$.

The given Fischer projection depicts tartaric acid. Observing the structure, there is a distinct plane of symmetry (indicated by the dashed line) dividing the molecule into two identical halves that are mirror images of each other.

Because of this internal plane of symmetry, the molecule is achiral and optically inactive despite having chiral centers. This specific isomer is known as meso-tartaric acid.

When an alkyl halide is treated with a very bulky, sterically hindered base such as potassium tert-butoxide ( $𝑡-{\mathrm{BuO}}^{-}{K}^{+}$), elimination proceeds via an E2 mechanism. Due to massive steric hindrance, the bulky $𝑡-{\mathrm{BuO}}^{-}$base struggles to extract a proton from the more substituted, internal carbon $\left({\mathrm{CH}}_2\right)$. Instead, it preferentially extracts a much more accessible proton from the terminal methyl $\left({\mathrm{CH}}_3\right)$ group. This favors the formation of the less substituted alkene, yielding 2-methyl-1-butene as the major product (known as the Hofmann product), rather than the thermodynamically more stable Zaitsev product.

The reaction sequence is explained below.

1. Formation of $𝑨$ Toluene $\left(C_6{H}_5{\mathrm{CH}}_3\right)$ is treated with strong oxidising agents ${\mathrm{KMnO}}_4,\mathrm{KOH},\Delta$. The alkyl side chain is completely oxidised to form benzoic acid $\left(C_6{H}_5\mathrm{COOH}\right)$. So, $𝐴$ is benzoic acid.

2. Formation of $𝑩$ Benzoic acid reacts with thionyl chloride $\left({\mathrm{SOCl}}_2\right)$ to replace the -OH group with -Cl , forming benzoyl chloride $\left(C_6{H}_5\mathrm{COCl}\right)$. So, $𝐵$ is benzoyl chloride.

3. Formation of $𝐂$ Benzoyl chloride is subjected to Rosenmund reduction $(H_2,\mathrm{Pd}/{\mathrm{BaSO}}_4)$. The poisoned catalyst selectively reduces acid chlorides to aldehydes, preventing over-reduction to an alcohol. This yields benzaldehyde $\left(C_6{H}_5\mathrm{CHO}\right)$.

The reaction is an intramolecular aldol condensation of a diketone (2,6-heptanedione).

The base abstracts an alpha-proton. Extracting a proton from the terminal methyl group (C1) yields an

enolate that can nucleophilically attack the C6 carbonyl carbon. This forms a stable, unstrained 6 -membered ring.

Following the attack, a cyclic $𝛽$-hydroxy ketone is formed. Upon heating, it dehydrates to form an $𝛼$, $𝛽$-unsaturated cyclic ketone.

The reaction sequence is shown below.

$\begin{array}{cc}& {\mathrm{CH}}_3-O-C{\left({\mathrm{CH}}_3\right)}_3+\mathrm{HI}⟶{\mathrm{CH}}_3\mathrm{OH}+{\left({\mathrm{CH}}_3\right)}_{3}C-I\end{array}$

The cleavage of an ether with cold, concentrated hydroiodic acid (HI) involves protonation of the ether oxygen followed by nucleophilic attack by the iodide ion $\left(I^{-}\right)$.

In tert-butyl methyl ether, one of the alkyl groups is tertiary. Because tertiary carbocations are extremely stable, the reaction proceeds via an $S_{N}1$ mechanism.

1. Step 1 The ether oxygen is protonated by HI.

2. Step 2 The C-O bond attached to the tert-butyl group breaks spontaneously to form a highly stable tert-butyl carbocation and methanol $\left({\mathrm{CH}}_3\mathrm{OH}\right)$ as the leaving group.

3. Step 3 The iodide ion $\left(I^{-}\right)$quickly attacks the stable tert-butyl carbocation. Consequently, the final products are tert-butyl iodide and methanol.

Hinsberg's reagent is used for distinguishing primary, secondary, and tertiary amines based on the solubility of their sulphonamide derivatives in alkali.

Chemically, Hinsberg's reagent is strictly benzene sulphonyl chloride $\left(C_6{H}_5{\mathrm{SO}}_2\mathrm{Cl}\right)$, shown in the structure above.

In Lassaigne's test, if an organic compound contains both nitrogen and sulphur, fusing it with sodium yields sodium thiocyanate (NaSCN).

$\mathrm{Na}+C+N+S⟶\mathrm{NaSCN}$

When iron(III) ions ( ${\mathrm{Fe}}^{3+}$ ) from ${\mathrm{FeCl}}_3$ are added to the extract, it forms a characteristic blood-red coloured coordination complex, iron(III) thiocyanate.

${\mathrm{Fe}}^{3+}+{\mathrm{SCN}}^{-}⟶\left[\mathrm{Fe}\right(\mathrm{SCN}){]}^{2+}$

Among the listed options, PHBV (Poly $𝛽$-hydroxybutyrate-co- $𝛽$-hydroxyvalerate) is a well-known biodegradable aliphatic polyester, often used in specialty packaging and orthopedic devices. Buna-N, nylon-6,6, and teflon are synthetic polymers that are essentially non-biodegradable under normal environmental conditions.

The alpha-helix ( $𝛼$-helix) is a common secondary structure of proteins. It represents a right-handed coiled conformation that is highly stable. This stability is predominantly maintained by extensive hydrogen bonds forming between the carbonyl oxygen $(-C=O)$ of one amino acid residue and the amide hydrogen $(-\mathrm{NH})$ of an amino acid residue situated four positions further down the polypeptide chain.

Sucralose is a popular artificial, zero-calorie sweetening agent. Structurally, it is synthesised by the selective chlorination of sucrose, where three hydroxyl groups are replaced with chlorine atoms, making it a trichloro derivative.

Photochemical smog (often referred to as Los Angeles smog) is a type of air pollution primarily arising from vehicular emissions.

It is driven by sunlight-induced photochemical reactions involving primary pollutants, specifically unburnt hydrocarbons and nitrogen oxides ( ${\mathrm{NO}}_{𝑥}$ ), which then generate secondary pollutants like ozone $\left(O_3\right)$ and Peroxyacetyl Nitrate (PAN).

The half-reaction for the reduction of dichromate $\left(K_2{\mathrm{Cr}}_2{O}_7\right)$ in an acidic medium is

${\mathrm{Cr}}_2{O}_{2-}^7+14H^{+}+6{𝑒}^{-}⟶2{\mathrm{Cr}}^{3+}+7H_{2}O$

This balanced equation indicates that the complete reduction of 1 mole of dichromate ions requires 6 moles of electrons.

For 0.1 moles of $K_2{\mathrm{Cr}}_2{O}_7$, the number of moles of electrons transferred is $0.1\times 6=0.6$ moles.

The Freundlich adsorption isotherm is mathematically expressed as:

$\frac{𝑥}{𝑚}=𝑘\cdot {𝑝}^{\frac{1}{𝑛}}\text{ (where, }\frac{𝑥}{𝑚}\text{ is the extent of adsorption). }$

At extremely high pressures, the available surface area of the adsorbent becomes fully saturated. Any further increase in pressure does not increase adsorption. At this saturation limit, the value of $\frac{1}{𝑛}$ approaches zero $(\frac{1}{𝑛}=0)$.

Therefore, $\frac{𝑥}{𝑚}\propto {𝑝}^{\circ }$, meaning the extent of adsorption becomes independent of pressure $𝑝$.

The relationship between equilibrium constants ${𝐾}_{𝑝}$ and ${𝐾}_{𝐶}$ is given by the formula ${𝐾}_{𝑝}={𝐾}_{𝐶}(𝑅𝑇{)}^{\Delta 𝑛𝑔}$.

For the Haber process reaction:

$N_2\left(𝑔\right)+3H_2\left(𝑔\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(𝑔\right)$

$\Delta {𝑛}_{𝑔}=$ (moles of gaseous products) - (moles of gaseous reactants) $=2-(1+3)=-2$

Substituting this value yields: ${𝐾}_{𝑝}={𝐾}_{𝐶}(𝑅𝑇{)}^{-2}$.

Resonance energy reflects the stability gained by a molecule due to electron delocalisation. Typically, fused aromatic ring systems display higher total resonance energy than isolated single rings or heterocyclic 5-membered rings.

Benzene has a single ring and a resonance energy of $\sim 36\mathrm{kcal}/\mathrm{mol}$.

Naphthalene consists of two fused benzene rings. With more extended conjugation across the $10𝜋$ electrons, its total resonance energy is ~61 kcal/mol.

Pyrrole and furan are heterocyclic rings and have lower resonance energies (approx. 21 and 16 kcal/mol respectively) compared to pure benzene.

Therefore, naphthalene has the maximum total resonance energy among the options.

Let's rigorously analyze the Lewis structure of the triiodide ion $\left(I_{-}^3\right)$.

The central iodine atom has 7 valence electrons, plus 1 extra electron for the overall negative charge, making 8. It uses 2 electrons to form single covalent bonds with the two terminal iodine atoms. The remaining 6 electrons form 3 lone pairs on the central atom(yielding an $𝑠{𝑝}^3𝑑$-hybridisation with a linear geometry). Each of the two terminal iodine atoms holds 3 lone pairs in order to complete their octets. Total lone pairs $=3($ central $)+3($ terminal 1$)+3$ $($ terminal 2$)=9$ lone pairs.

Victor Meyer's test is a classical chemical test used to distinguish between primary, secondary and tertiary alcohols based on color changes after a series of reactions with $P/I_2,{\mathrm{AgNO}}_2$, and ${\mathrm{HNO}}_2$, followed by alkalisation (NaOH).

Primary alcohols $\left(1^{\circ }\right)$ Undergo the sequence of reaction to form nitrolic acid.

$𝑅-{\mathrm{CH}}_2\mathrm{OH}⟶𝑅-{\mathrm{CH}}_{2}I\to 𝑅-{\mathrm{CH}}_2{\mathrm{NO}}_2⟶$

Secondary alcohols $\left(2^{\circ }\right)$ Form a pseudonitrole.

${𝑅}_2\mathrm{CHOH}⟶{𝑅}_2\mathrm{CHI}⟶{𝑅}_2{\mathrm{CHNO}}_2$

Tertiary alcohols $\left(3^{\circ }\right)$ Form a tertiary nitro compound that does not react with nitrous acid $\left({\mathrm{HNO}}_2\right)$. ${𝑅}_3\mathrm{COH}\to {𝑅}_3\mathrm{CI}\to {𝑅}_3{\mathrm{CNO}}_2\to$ No reaction with ${\mathrm{HNO}}_2+\mathrm{NaOH}\to$ Colourless

$\text{ The correct sequence is SPRQ. }$

$\text{ The correct sequence is QPSR. }$

'Denouement' means the outcome of a situation, when something is decided or made clear. So, 'outcome' is the correct answer.

'Gauche' means lacking social experience or grace; socially awkward or not tactful. So, 'Tactless' would be its correct similar meaning word.

'Reprimand' means a formal expression of disapproval. So, 'Appreciate' would be its correct antonym.

'Impertinent' means not showing proper respect; rude. So, 'polite' would be its correct opposite meaning word.

A picture or a design made by putting together coloured pieces of glass or stones is called Mosaic.

The wrongly spelled word is 'temperture'. Its correct spelling in 'temperature'.

The suitable prefix for the word 'appear' is 'dis' and the word formed is 'disappear'.

$\text{ The given line is an example of personification. }$

As a Doctor works in a Hospital, similarly, a Teacher works in a School

Apple, Mango and Banana are all fruits, whereas Potato is a modified underground stem (vegetable). Therefore, Potato is the odd one out.

The given series follows the pattern of squares of consecutive natural numbers starting from 2 :

$\begin{array}{cc}& 2^2=4\\ & 3^2=9\\ & 4^2=16\\ & 5^2=25\\ & 6^2=36\\ & 7^2=49\end{array}$

The pattern is that each letter is shifted forward by 2 positions in the English alphabetical order. In the series

Similarly, $D(+2)\to F,O(+2)\to Q,G(+2)\to 𝐼$

As per the question, A is the brother of B and D is the sister of B . This establishes that $A,B$ and D are siblings. Since C is the father of $A,C$ must inherently be the father of all three siblings.

$\therefore 𝐶\text{ is the father of }𝐷\text{. }$

Let the starting point be O . The boy walks 5 km East, turns left (North) for 5 km and turns left (West) again for 5 km.

From the diagram, the distance from starting point O to final point $𝐶$ is parallel and equal to $𝐴𝐵$. Distance $\mathrm{OC}=5\mathrm{km}$.

The given expression is $15-3\times 5+\frac{2}{4}$

By replacing the symbols as per the given conditions ('-' becomes ' $𝑥$ ', ' $𝑥$ ' becomes ' ${/}^{\prime }$, '+' becomes '-', and ' ${/}^{\prime }$ becomes '+'), the new expression is

$=15\times \frac{3}{5}-2+4$

$\begin{array}{cc}& \text{ Applying the BODMAS rule: }\\ & \begin{array}{cc}\Rightarrow & 15\times \left(\frac{3}{5}\right)-2+4\\ \Rightarrow & 9-2+4\\ \Rightarrow & 7+4=11\end{array}\end{array}$

As per the statements, we draw the possible Venn diagrams

Conclusions

I. (x)                 II. (x)

It is entirely possible that the 'Sweet' circle intersects 'Red Things' but does not intersect the 'Apples' sub-category. Thus, neither conclusion definitely follows

As per the given data:

Total number of students $=30$

Rohan's rank from top $=10$

Rank from bottom $=($ Total students - Rank from top $)+1$

$\Rightarrow (30-10)+1=20+1=21$

As per the question, Dogs and Cats are distinct species but both fall entirely under the Animal category:

In the given figure series, the dot shifts exactly one quadrant in the clockwise direction in each successive step.

Therefore, in the next step, the dot will naturally appear in the bottom-left quadrant.

Pattern of the given shape analogy is as follows:

Similarly, adding +1 side to the initial figure (triangle) yields a square shape.

As per the options provided:

Except option (d), all other figures are divided into exactly two equal sections by a single straight line. Option (d) is divided into 4 parts.

In a simple square divided by its two main diagonals:

Number of individual small triangles $=4$ (labeled 1, 2, 3, 4)

Number of large triangles formed by combining two adjacent small ones $=4(1+2,2+3,3+4,4+1)$

Total triangles $=4+4=8$.

It is clear that answer figure (b) completes the original figure, which looks like as shown in the adjacent figure.

Hence, alternative (b) is the correct answer.

As per the question, the mirror image of the given arrow is

The horizontally reflected arrow will correctly point towards the North-West direction.

Based on the statement, the Venn diagram relationships are:

The specific region that falls strictly inside the square (Actors) and Circle (Singers) but strictly outside the Triangle (Dancers) contains the number 4.

The figure (a) [the ' $𝑇$ ' shape] is hidden in option

(b)'s figure as its part without rotation.

The alphabetical letters are coded by assigning their positional numerical values: $A=1,B=2,C=3,\dots$ $Z=26$.

For the word 'BAT':

$\begin{array}{cc}B& =2\\ A& =1\\ T& =20\\ \therefore \mathrm{BAT}& =2+1+20\\ & =23\end{array}$

As per the rules of a standard dice, the sum of numbers on any two opposite faces is exactly 7.

Therefore, on the face opposite to 2 , the number 5 will appear (since $7-2=5$ ).