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ISRO Scientist/Engineer Electronics 2025 question paper

All 95 questions of ISRO's written test for Scientist/Engineer 'SC' (Electronics), held on 26 October 2025, with ISRO's answer key. ISRO excluded Q.44, Q.49 and Q.82 from evaluation.

Answers from ISRO's final answer key of 15 May 2026.

The question paper (PDF, ISRO)ISRO’s answer key (PDF)

questions · 100 marks: 95questions · 100 marks
Part A: Electronics: 80Part A: Electronics
Part B: aptitude: 15Part B: aptitude

How the paper is marked

Part A (Q.1 to Q.80): 1 mark for a correct answer, 0 for no answer, minus 0.33 for a wrong answer.
Part B (Q.81 to Q.95): 1 or 2 marks as printed with each question, 20 marks in all, no negative marking.

Source: the instructions printed in the question paper.

ISRO EC 2025 answer key

ISRO's answer key for ISRO Scientist/Engineer Electronics 2025
Question	Part	Marks	Answer
Q.1	A	1	B
Q.2	A	1	C
Q.3	A	1	C
Q.4	A	1	D
Q.5	A	1	C
Q.6	A	1	D
Q.7	A	1	C
Q.8	A	1	D
Q.9	A	1	C
Q.10	A	1	A
Q.11	A	1	B
Q.12	A	1	B
Q.13	A	1	D
Q.14	A	1	B
Q.15	A	1	C
Q.16	A	1	A
Q.17	A	1	B
Q.18	A	1	C
Q.19	A	1	C
Q.20	A	1	A
Q.21	A	1	A
Q.22	A	1	C
Q.23	A	1	C
Q.24	A	1	B
Q.25	A	1	B
Q.26	A	1	B
Q.27	A	1	D
Q.28	A	1	B
Q.29	A	1	C
Q.30	A	1	A
Q.31	A	1	D
Q.32	A	1	D
Q.33	A	1	C
Q.34	A	1	C
Q.35	A	1	A
Q.36	A	1	C
Q.37	A	1	D
Q.38	A	1	C
Q.39	A	1	B
Q.40	A	1	C
Q.41	A	1	C
Q.42	A	1	A
Q.43	A	1	B
Q.44	A	1	Excluded
Q.45	A	1	C
Q.46	A	1	A
Q.47	A	1	B
Q.48	A	1	B
Q.49	A	1	Excluded
Q.50	A	1	D
Q.51	A	1	B
Q.52	A	1	C
Q.53	A	1	D
Q.54	A	1	A
Q.55	A	1	A
Q.56	A	1	D
Q.57	A	1	A
Q.58	A	1	B
Q.59	A	1	C
Q.60	A	1	B
Q.61	A	1	A
Q.62	A	1	B
Q.63	A	1	C
Q.64	A	1	C
Q.65	A	1	B
Q.66	A	1	C
Q.67	A	1	C
Q.68	A	1	A
Q.69	A	1	A
Q.70	A	1	A
Q.71	A	1	A
Q.72	A	1	A
Q.73	A	1	C
Q.74	A	1	C
Q.75	A	1	C
Q.76	A	1	C
Q.77	A	1	C
Q.78	A	1	A
Q.79	A	1	B
Q.80	A	1	B
Q.81	B	1	D
Q.82	B	1	Excluded
Q.83	B	2	B
Q.84	B	1	C
Q.85	B	1	B
Q.86	B	2	B
Q.87	B	1	D
Q.88	B	1	C
Q.89	B	2	D
Q.90	B	1	C
Q.91	B	1	B
Q.92	B	2	D
Q.93	B	1	D
Q.94	B	1	C
Q.95	B	2	A

ISRO excluded Q.44, Q.49 and Q.82 from evaluation.

Part A: Electronics Q.1 to Q.80

Q.11 markOfficial answer: B
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Signals and Systems › Laplace
Transform each term, combine into a single rational function, and read poles from the denominator and zeros from the numerator after simplifying.
Full solution
Q.21 markOfficial answer: C
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Networks (Electric Circuits) › AC, power & resonance
In an L-match the shunt element goes across the larger resistance and the series element on the smaller side, with $Q = R_{high} / R_{low} - 1$ . For a high-pass match, use a series capacitor and a shunt inductor.
Full solution
Q.31 markOfficial answer: C
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Earlier, ISRO's key gave B.
Analog Electronics › Amplifiers
Miller's theorem: a bridging impedance $Z$ appears at the input as $Z / (1 - A)$ , so a bridging capacitor $C$ appears as $(1 - A) C$ , where $A = V_{o} / V_{i}$ is signed.
Full solution
Q.41 markOfficial answer: D
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Earlier, ISRO's key gave C.
Signals and Systems › Signals basics
A periodic signal must satisfy $x (t + T) = x (t)$ for all time. Multiplying by a unit step, or adding a single isolated sample, breaks periodicity. An impulse train that repeats every 4 samples, or a difference of two such trains, is periodic with period 4.
The question and its official answer
Q.51 markOfficial answer: C
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Electromagnetics › Antennas
EIRP is radiated power multiplied by antenna gain over isotropic; 3 dBi is a power factor of about 2, so the EIRP is double the radiated power.
Full solution
Q.61 markOfficial answer: D
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Q.71 markOfficial answer: C
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Signals and Systems › Fourier transform
A sinc in time is a rectangle in frequency: the spectrum has finite support (band limited), while the time signal extends over all time (not time limited). A non-zero signal cannot be limited in both domains.
Full solution
Q.81 markOfficial answer: D
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Q.91 markOfficial answer: C
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Analog Electronics › Op-amps
The virtual short $v_{+} \approx v_{-}$ comes from $v_{+} - v_{-} = v_{o} / A_{O L}$ . It holds only when the output is finite (linear region, not saturated), which negative feedback ensures, together with a very high gain.
Full solution
Q.101 markOfficial answer: A
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Q.111 markOfficial answer: B
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Electromagnetics › Transmission lines
On a lossless line, $v_{p} = 1/ L C$ using the per-metre inductance and capacitance. Length and frequency are distractors.
Full solution
Q.121 markOfficial answer: B
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Q.131 markOfficial answer: D
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Signals and Systems › Laplace
Factorise the denominator into real poles, expand in partial fractions using the residue at each pole, then use $\frac{1}{s + a} \leftrightarrow e^{- a t} u (t)$ .
Full solution
Q.141 markOfficial answer: B
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Q.151 markOfficial answer: C
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Q.161 markOfficial answer: A
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Q.171 markOfficial answer: B
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Signals and Systems › z-transform
A delay of $k$ samples in time multiplies the Z-transform by $z^{- k}$ , so $δ [n - k] \leftrightarrow z^{- k}$ . The sum $\sum x [n] z^{- n}$ keeps just the single term at $n = k$ .
Full solution
Q.181 markOfficial answer: C
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Engineering Mathematics › Probability & statistics
Memoryless means $P (X > s + t ∣ X > s) = P (X > t)$ . This forces the tail to satisfy $S (s + t) = S (s) S (t)$ , whose only continuous solution is $e^{- λ x}$ , the exponential distribution (the geometric distribution is the discrete analogue).
Full solution
Q.191 markOfficial answer: C
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Electromagnetics › Transmission lines
A line one wavelength long (or any whole number of half-wavelengths) repeats its load impedance, so only the quarter-wave section transforms it, by $Z_{in} = Z_{0}^{2} / Z_{L}$ .
Full solution
Q.201 markOfficial answer: A
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Signals and Systems › LTI & convolution
Convolution with $δ [n - k]$ is a delay by $k$ samples, so an impulse response made of impulses gives an output made of delayed copies of the input.
Full solution
Q.211 markOfficial answer: A
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Engineering Mathematics › Probability & statistics
Gaps between arrivals of a Poisson process are exponential with the same rate, so $P (T > t) = e^{- λ t}$ , provided $λ$ and $t$ are in the same time unit.
Full solution
Q.221 markOfficial answer: C
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Electromagnetics › Antennas
Add the dB terms: received power = EIRP (transmit power plus gain) minus the spreading loss $10 lo g_{10} (4 π R^{2})$ plus $10 lo g_{10} A_{e}$ . Keep $R$ in metres and the units consistent (dBW, dB relative to $1 m^{2}$ ).
Full solution
Q.231 markOfficial answer: C
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Q.241 markOfficial answer: B
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Digital Electronics › Combinational
A half subtractor has $D = A \oplus B$ and borrow $X = \overline{A} B$ , because a borrow occurs only for $0 - 1$ . Build the truth table first and read the expressions from it.
Full solution
Q.251 markOfficial answer: B
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Electromagnetics › Waveguides & fibres
TE wave impedance is $ω μ / β$ . Because $β = k 1 - (f_{c} / f)^{2}$ is smaller than $k$ , the impedance is larger than $η$ and grows without limit at cutoff. TM is the reverse: $Z_{T M} = η 1 - (f_{c} / f)^{2}$ .
Full solution
Q.261 markOfficial answer: B
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Digital Electronics › ADC & DAC
An ideal ADC gains about 6.02 dB of SNR for every extra bit ( $SNR = 6.02 N + 1.76 dB$ ), so adding 2 bits adds about 12 dB.
Full solution
Q.271 markOfficial answer: D
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Q.281 markOfficial answer: B
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Electromagnetics › Antennas
Friis equation: $P_{r} = P_{t} G_{t} G_{r} (λ /4 π R)^{2}$ . Only the $λ^{2}$ factor is a direct (positive) power-law dependence among the options; distance and frequency appear in the denominator, and transmitted power appears to the first power.
Full solution
Q.291 markOfficial answer: C
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Electronic Devices › Carrier transport
Mobility is $q τ / m^{*}$ . At and above room temperature, phonon (lattice) scattering dominates and grows with $T$ , giving $μ \propto T^{- 3/2}$ .
Full solution
Q.301 markOfficial answer: A
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Q.311 markOfficial answer: D
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Control Systems › Routh–Hurwitz
Build the Routh array and require every first-column entry to be positive. The only entries that depend on $K$ are $4 - 2 K$ and $K$ , which give $0 < K < 2$ .
Full solution
Q.321 markOfficial answer: D
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Engineering Mathematics › Linear algebra
A similarity transformation $B = P^{- 1} A P$ preserves the characteristic polynomial, and with it the eigenvalues, trace (sum of eigenvalues) and determinant (product of eigenvalues).
Full solution
Q.331 markOfficial answer: C
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Q.341 markOfficial answer: C
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Networks (Electric Circuits) › Laws & node/mesh
Write KCL at the two unknown nodes. The ideal voltage source fixes the left node at $75 V$ , and the ideal current source simply forces $6 A$ out of the right node whatever the voltage across it.
Full solution
Q.351 markOfficial answer: A
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Engineering Mathematics › Differential equations
For $u_{t} = c u_{xx}$ with $u = 0$ at $x = 0, π$ , the mode $sin (n x)$ decays as $e^{- c n^{2} t}$ . A single-mode initial condition stays a single mode.
Full solution
Q.361 markOfficial answer: C
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Control Systems › Compensators & PID
Lead and lag compensators share one form, a gain with a single zero and a single pole: $K (1 + s T_{1}) / (1 + s T_{2})$ . Lead when $T_{1} > T_{2}$ , lag when $T_{1} < T_{2}$ .
Full solution
Q.371 markOfficial answer: D
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Networks (Electric Circuits) › Transients
A capacitor voltage is continuous, so take the initial value from the old steady state (with its sign from the polarity marks) and the final value and time constant from the new circuit alone: open the capacitor for the final value, and switch off the sources (current source open) to find the resistance for the time constant.
Full solution
Q.381 markOfficial answer: C
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Analog Electronics › Amplifiers
Output resistance of a MOSFET in saturation is $r_{o} = 1/ (λ I_{D})$ , so it is inversely proportional to $λ$ : a larger $λ$ gives a steeper $I_{D}$ - $V_{D S}$ curve and a lower $r_{o}$ .
Full solution
Q.391 markOfficial answer: B
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Q.401 markOfficial answer: C
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Digital Electronics › Sequential
A Mealy detector must remember the count of recent 0s (none, one, two or more) and whether it is inside a qualifying run of 1s. Merge states with identical future behaviour, then check that the rest are pairwise distinguishable by some input sequence.
Full solution
Q.411 markOfficial answer: C
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Digital Electronics › Sequential
$MTBF = e^{t_{r} / τ} / (T_{0} f_{c l k} f_{d a t a})$ : anything that adds more sampling events or data edges per second lowers MTBF, while a smaller window or more resolution time raises it.
Full solution
Q.421 markOfficial answer: A
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Signals and Systems › Fourier transform
The transform of a rectangular pulse of width $T$ and height 1 is $T \frac{sin ( ω T /2 )}{ω T /2}$ : its value at $ω = 0$ is the pulse area, and its first zero is at $ω = 2 π / T$ . Here $T = 2$ , so the area is 2 and the first zero is at $ω = π$ .
Full solution
Q.431 markOfficial answer: B
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Analog Electronics › Diode circuits
For ideal diodes, assume a set of on/off states, solve the circuit, and keep the assumption only if every conducting diode carries positive current and every off diode has reverse voltage across it.
Full solution
Q.441 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave B.
Q.451 markOfficial answer: C
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Q.461 markOfficial answer: A
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Analog Electronics › Op-amps
To find which op-amp input is inverting, raise the output and follow the loop through the circuit. The input that rises with the output must be the inverting one. Here a higher output lowers $V_{GS}$ , so the transistor current falls and the drain node (terminal B) rises.
Full solution
Q.471 markOfficial answer: B
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Control Systems › Block diagrams & SFG
Build the signal flow graph in alternating form. Each current is a conductance times a voltage difference, so its branches have gain $1/ R$ . Each node voltage is a resistance times a current, so its branches have gain $R$ .
Full solution
Q.481 markOfficial answer: B
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Electromagnetics › Antennas
Gain includes the ohmic losses and directivity does not, so $G = η D$ with $η = R_{r} / (R_{r} + R_{L})$ . Find $D$ as a ratio first, then convert to dB.
Full solution
Q.491 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave B.
Q.501 markOfficial answer: D
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Control Systems › Compensators & PID
PD control adds a zero: phase lead and damping improve stability and speed, but the gain rises with frequency (noise sensitivity), and it leaves the DC gain, and so the steady-state error, unchanged.
Full solution
Q.511 markOfficial answer: B
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Electromagnetics › Maxwell's equations
Across a boundary with no surface current, tangential $H$ is continuous (Ampere's law) and normal $B$ is continuous (Gauss's law for magnetism), so normal $H$ jumps by the ratio of permeabilities.
Full solution
Q.521 markOfficial answer: C
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Analog Electronics › Biasing
In the active region, find $I_{E}$ from the emitter loop, take $I_{C} = α I_{E}$ , then apply KVL around the collector loop to get the collector voltage relative to the base.
Full solution
Q.531 markOfficial answer: D
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Control Systems › Time response
With non-unity feedback, define the error as $R - C$ , not as the summing-junction output. Get $e_{ss}$ from the final value theorem, then apply $S = \frac{u}{e _{ss}} \frac{d e _{ss}}{d u}$ .
Full solution
Q.541 markOfficial answer: A
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Electromagnetics › Plane waves
The loss tangent is the ratio of conduction current to displacement current, $tan δ = σ / (ω ε^{'})$ . A very large value means conduction dominates, which is conductor behaviour.
Full solution
Q.551 markOfficial answer: A
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Analog Electronics › Op-amps
With negative feedback and the non-inverting input grounded, the inverting input is a virtual ground and draws no current, so the input current passes entirely through the feedback resistor. The voltage it drops fixes the junction voltage, and Ohm's law on the shunt resistor gives the third current.
Full solution
Q.561 markOfficial answer: D
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Control Systems › Bode & margins
Read the corner frequencies and slope changes off the asymptotic Bode plot: each $- 20 dB/decade$ step is a simple pole, and the gain $K$ comes from the low-frequency segment, here $K / ω = 100$ at $ω = 1$ .
Full solution
Q.571 markOfficial answer: A
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Q.581 markOfficial answer: B
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Analog Electronics › Filters & waveforms
An op-amp with no feedback is a comparator: the output goes high when the non-inverting input exceeds the inverting input. Find the reference from the resistor divider, then check which output polarity forward-biases the LED.
Full solution
Q.591 markOfficial answer: C
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Analog Electronics › Feedback & oscillators
Identify the topology by two tests. At the output, set $V_{o} = 0$ : if the feedback vanishes, the sampling is of voltage (shunt). At the input, ask whether the feedback is subtracted as a voltage in series with the source (series mixing) or as a current at a node (shunt mixing). Here the feedback is a voltage in series and it is sampled from $V_{o}$ , giving series-shunt.
Full solution
Q.601 markOfficial answer: B
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Signals and Systems › Sampling
Nyquist rate = 2 x the highest frequency present in the signal. Convert each angular frequency with $f = ω /2 π$ and take the largest.
Full solution
Q.611 markOfficial answer: A
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Earlier, ISRO's key gave C.
Q.621 markOfficial answer: B
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Q.631 markOfficial answer: C
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Q.641 markOfficial answer: C
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Electronic Devices › Carrier transport
Diffusion current density is $J = \pm q D \frac{d ( carrier concentration )}{d x}$ , so it grows with the concentration gradient and vanishes when the profile is uniform.
Full solution
Q.651 markOfficial answer: B
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Networks (Electric Circuits) › AC, power & resonance
For a series R-L circuit, the power factor is $cos ϕ = R /∣ Z ∣$ , where $tan ϕ = X_{L} / R$ . With $X_{L} = R$ the phase angle is $4 5^{\circ}$ . The current lags the voltage because the load is inductive.
Full solution
Q.661 markOfficial answer: C
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Digital Electronics › Logic families
In the transition region $V_{T n} < V_{in} < V_{D D} - ∣ V_{T p} ∣$ , so both the NMOS and the PMOS are on at once. Near $V_{M}$ both are in saturation, which gives the high gain and the short-circuit current peak.
Full solution
Q.671 markOfficial answer: C
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Q.681 markOfficial answer: A
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Electronic Devices › Semiconductors
In an N-type semiconductor $n \approx N_{D} = N_{C} e^{- (E_{C} - E_{F}) / k T}$ , so $E_{F}$ sits below $E_{C}$ by $k T ln (N_{C} / N_{D})$ .
Full solution
Q.691 markOfficial answer: A
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Q.701 markOfficial answer: A
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Electronic Devices › Semiconductors
The band gap of a semiconductor is largest at $0 K$ : about $0.74 eV$ for germanium (versus $0.67 eV$ at room temperature) and about $1.17 eV$ for silicon (versus $1.12 eV$ at room temperature).
Full solution
Q.711 markOfficial answer: A
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Signals and Systems › z-transform
A first-order pole factor $\frac{1}{1 - a e ^{- j ω}}$ is the DTFT of the causal decaying exponential $a^{n} u [n]$ (for $∣ a ∣ < 1$ ).
Full solution
Q.721 markOfficial answer: A
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Electronic Devices › MOS devices
$V_{P}$ is the gate-to-source voltage at which the channel is fully depleted and $I_{D} = 0$ , so that $I_{D} = I_{D S S} (1 - V_{GS} / V_{P})^{2}$ vanishes at $V_{GS} = V_{P}$ .
Full solution
Q.731 markOfficial answer: C
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Signals and Systems › Laplace
Poles are the roots of the denominator. A negative discriminant gives a complex-conjugate pair, and the sign of the real part ( $- ζ ω_{n}$ ) tells you stability.
Full solution
Q.741 markOfficial answer: C
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Analog Electronics › Filters & waveforms
A Schmitt trigger has two thresholds, $V_{U T}$ and $V_{L T}$ . Its output depends on the input and on its previous state, through the hysteresis $V_{H} = V_{U T} - V_{L T}$ .
Full solution
Q.751 markOfficial answer: C
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Signals and Systems › LTI & convolution
An LTI system is causal exactly when $h [n] = 0$ for all $n < 0$ . Find where each shifted impulse or step is nonzero. A term like $δ [n + 1]$ sits at $n = - 1$ , which is in the past, so it uses a future input.
Full solution
Q.761 markOfficial answer: C
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Digital Electronics › Processors & memory
Chips sharing the same address bus (and select) give word-size expansion: locations stay the same, data width adds up. Chips on different address ranges give depth expansion: locations add up, width stays the same.
Full solution
Q.771 markOfficial answer: C
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Q.781 markOfficial answer: A
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Digital Electronics › Sequential
Stages of a ripple counter change one after another, so the total settling time grows as $n t_{p d}$ and caps the clock frequency at about $1/ (n t_{p d})$ .
Full solution
Q.791 markOfficial answer: B
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Networks (Electric Circuits) › AC, power & resonance
In a parallel RLC circuit fed through a series resistor, the source resistance loads the tank: Q and bandwidth use the series resistor in parallel with the shunt resistor, while the resonant frequency depends only on L and C.
Full solution
Q.801 markOfficial answer: B
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Networks (Electric Circuits) › Transients
At $t = 0^{+}$ an inductor with zero initial current behaves like an open circuit: its current stays at zero, so the resistor drops nothing and the whole source voltage appears across the inductor.
Full solution

Part B: Aptitude Q.81 to Q.95

Q.811 markOfficial answer: D
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General Aptitude › Numerical
Count odd days (days modulo 7) from a known reference date: an ordinary year adds 1 odd day, a leap year adds 2, and a whole number of weeks adds 0.
Full solution
Q.821 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave B.
Q.832 marksOfficial answer: B
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Earlier, ISRO's key gave A.
General Aptitude › Reasoning
A floor number tells you the least number of floors its building has. A 7th floor means at least 7 floors, which is more than 3, so the rule applies and that building has a lift. A conditional "if more than 3, then lift" says nothing about buildings with 3 or fewer floors.
Full solution
Q.841 markOfficial answer: C
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General Aptitude › Geometry
Volume scales as the cube of the linear scale factor: doubling every side multiplies the volume by $2^{3} = 8$ .
Full solution
Q.851 markOfficial answer: B
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General Aptitude › Numerical
When only one component is added, the other component's quantity stays fixed. Use it as the anchor and set up the new ratio against it.
Full solution
Q.862 marksOfficial answer: B
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General Aptitude › Statistics & probability
Discrete variables take countable values and have a probability mass function. Poisson, binomial and geometric are the standard discrete laws. Gaussian, Rayleigh, exponential and uniform are continuous.
Full solution
Q.871 markOfficial answer: D
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Earlier, ISRO's key gave C.
General Aptitude › Numerical
Subtract the sum of the people in a given group from the total to isolate the one who is left out, then back-substitute into the other equations.
Full solution
Q.881 markOfficial answer: C
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General Aptitude › Reasoning
When neither conclusion follows by itself but the two are contradictories (a particular affirmative and a universal negative on the same two terms), one of them must be true, so the answer is 'Either I or II follows'.
Full solution
Q.892 marksOfficial answer: D
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General Aptitude › Reasoning
A conclusion follows only if the statements force it. Statements about women teachers say nothing positive about other athletes, so claims that other athletes can play are not forced.
Full solution
Q.901 markOfficial answer: C
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General Aptitude › Reasoning
When a number series grows quickly, take successive differences, and if they are not constant take differences of the differences. Here the differences are $2^{k} + 1$ , so the next one is $17$ .
Full solution
Q.911 markOfficial answer: B
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General Aptitude › Spatial
A mirror image does two things at once: it reverses the order of the characters and it flips every character left to right. Check the first and last characters of each option first, then check the glyph shapes.
Full solution
Q.922 marksOfficial answer: D
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General Aptitude › Numerical
Each circle's total includes its overlap regions, so subtract the labelled overlaps to find the unlabelled "only" regions. Then add only the regions lying outside the excluded circle.
Full solution
Q.931 markOfficial answer: D
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General Aptitude › Spatial
In a pattern grid, follow each symbol on its own. Here the circle shifts one square right and the star shifts one square up in every box along a row, and the same two shifts hold in all three rows.
Full solution
Q.941 markOfficial answer: C
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Earlier, ISRO's key gave A.
General Aptitude › Spatial
Separate the two marks. The black square follows a chessboard parity rule, and the crossed square repeats between cells a knight's move apart.
Full solution
Q.952 marksOfficial answer: A
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General Aptitude › Spatial
Track two things separately: how the shape grows (one square added along a fixed path) and how the colouring behaves (alternating along the path, flipping its starting colour each figure).
Full solution

WHERE THE MARKS CAME FROM

Marks by subject in ISRO Scientist/Engineer Electronics 2025
Subject	Qs	Marks
General Aptitude	14	19
Signals and Systems	11	11
Analog Electronics	10	10
Electromagnetics	9	9
Digital Electronics	7	7
Control Systems	6	6
Networks (Electric Circuits)	6	6
Electronic Devices	5	5
Engineering Mathematics	4	4
Other questions	23	23

We sorted the questions into subjects ourselves; the exam paper does not label them.

Every ISRO paper ISRO Scientist/Engineer: who can apply ECE study material AEIE study material